Two important skills to have in mathematics are factorising quadratics and expanding brackets. They are opposite to each other and are used to simplify or expand expressions to or from the form:

where ab and c are constant coefficients and x is the unknown variable. a is said to be the coefficient of b the coefficient of x and c is the constant term.

Guide contents

The tabs of this guide will support you in quadratics factorising. The sections are organised as follows:

• Expanding Brackets - recognising quadratics and how to multiply them out
• Cases When a =/= 1 - how to process a quadratic equation when the coefficient of  is not one
• Cases When c = 0 - how to process a quadratic equation with no constant term
• Difference of Two Squares - noticing easy quadratics!
• Completing the Square - what to do when a quadratic will not factorise 'properly'

To use this guide, you need to be familiar with certain terminology:

• Coefficient - coefficients in quadratic equations are the numbers, or constants, which are placed before a variable such as x.
• Product - the product of two numbers is the result you get when you multiply them together. The product of 5 and 3 is 15, and the product of a and is ab.
• Parentheses - parentheses in mathematics are round brackets which surround expressions to group them together and to clarify orders of operations. 2(5 x 3)=2(15)=30, and 4(b)=4ab.

where ab and c are constant coefficients and x is what we are trying to solve for. a is known as the coefficient of b is known as the coefficient of x and c is the constant term.

Simple Expressions, When the Coefficients of x² and x is Zero

Recall the order of operations when dealing with brackets in mathematical equations. They are:

B - Brackets

I - Indices (or O - Operators)

D - Division

M - Multiplication

S - Subtraction

By following the order, complicated mathematical equations and expressions are easy to simplify and solve. For factorising quadratics, remember that anything inside of a bracket must be 'dealt with' first.

It is easy to expand the brackets (or parentheses) of a function when all the factors are known: we simply perform the operations inside the brackets first, then do the same with what is outside the brackets.

(16 - 4) + 3 = (12) + 3 = 12 + 3 = 15

16 - (4 + 3) = 16 - (7) = 16 - 7 = 9

16 + (-4 + 3) = 16 + (-1) = 16 - 1 = 15

You can see in the above examples that the location of the brackets and addition/subtraction signs are important, and the variations end up yielding varying results.

Some more examples are below:

2 x (5 + 3) = 2 x (8) = 2 x 8 = 16

(2 x 5) + 3 = (15) + 3 = 15 + 3 = 18

2 x (5 x 3) = 2 x (15) = 2 x 15 = 30

This holds true no matter how long your equation is.

20 + 7(103 x 0.39) - 180 = 20 + 7(40.17) - 180 = 20 + (281.19) - 180 = 20 + 281.19 - 180 = 121.19

Simple Expressions, When the Coefficient of x² is Zero

Performing the relevant operations inside or outside the parentheses is only slightly more difficult to do when one or more of the factors is unknown:

2(x 3) = 2(3x) = 6x

x(14 - 19) = x(-5) = -5x

8x + x(12x +1) - (2x - 4 - 1) = 8x + (12x² + x) - (2x - 5) = 12 + 7x +5

In this last example, we collected all the 'like' terms together - that is, we put all the coefficients of x² together, all the coefficients of x and the constant coefficients together, and simplified to make an expression which is both easier to read and to understand.

Having more than one unknown variable is no different either:

8(x + y) = 8x + 8y

-2x(a - bc) = -2xa + 2xbc

-(x + y) + x(4 + 3y) = -x - y + (4x + 3xy) = 3x - y + 3xy

In all of these examples, including the ones above, what is contained inside the parentheses is performed first, and anything outside the parentheses is performed afterwards. In the cases where one or more variable is unknown, it may be the case that what is contained inside may already be in its simplest form, and so there is nothing for us to do here - we can continue with the work which needs to be done outside the parentheses.

Expanding the Brackets Means Removing Them!

Observe the following example:

(x + 1)(x + 2) =  + 3x + 2

How did this happen? In the exact same way as what happened above - by multiplying out.

When working with brackets, we multiply each item in one set of brackets with each item in the other set in order. In other words:

(x + 1)(x + 2) = (x x x) + (x x 2) + (1 x x) + (1 x 2) =  + 1x + 2x + 2 =  + 3x + 2.

Step-By-Step Breakdown

Here, we have taken the first item in both brackets, which is both x in this case, and multiply them together. This is the first bracket, (x x x).

Then, we take the first item in the first bracket and multiply by the next item in the second bracket, which here is x and 2, (x x 2).

There are now no more items in the second bracket to consider, so we move on to the next stage, which is to take the second item in the first bracket and multiply it by the first item in the second bracket, which here is 1 and x, (1 x x).

Lastly, we take the second item in the second bracket and multiply it by the second item in the second bracket, which here is 1 and 2, (1 x 2).

There are no more items to consider, so there is no more to multiply together! We can now add up what we have:

(x x x) + (x x 2) + (1 x x) + (1 + 2)

To generalise this, we can write the following:

(x + m)(x + n) = + (n + m)x + mn

for constants m and n.

When dealing with parentheses inside of other parentheses, these inside-most ones need to be evaluated first before attempting any outside ones.

-(4x(a + b)) = -(4xa + 4xb) = -4xa - 4xb

x + y(1 + (xy + 2)) = x + y(3 + xy) = x + 3y + xy²

0.5(4(x + 1) - 2x(1 + y)) = 0.5(4x + 4 - 2x - 2xy) = 0.5(2x + 4 - 2xy) = x + 2 -xy

Remember that the order of operations matters! This ordering results in everyone being able to solve equations the same way.

What is Factorising?

Factorising is almost the opposite of expanding the brackets, as it involves removing the brackets of an expression and separating it into its factors. When we factorise a quadratic expression, we are rewriting it as an expression of multiplicative factors in its simplest form.

To factorise an expression, we need to take out the highest common factor:

14 + 12x = 2(7 + 6x)

x² - 3x = x(x - 3)

4x + 4y - (8z + 12) = 4(x + y) - 4(z + 3) = 4(x + y - 2z - 3)

The Greatest Common Factor

Recall that the greatest common factor between a set of numbers is the highest number which divides into each number of the set. What use does the greatest common factor play in quadratics? Well, when simplifying expressions, there may be more than one number in common which divides completely in your coefficients. If you choose one of the lesser common factors you will end up with an expression which can be simplified further, and you will end up having to do the factorising again! You might as well save yourself some work and choose the greatest common factor first time.

The expression

14x + 56

has the factors 2, 7 and 14 in common (that is, all of 2, 7 and 14 will divide into both 14 AND 56). If you choose to take 2 'out', your expression then becomes:

2(7x + 28)

Of course, then the inside of the brackets can be simplified further,  by taking 7 out:

2(7(x + 4))

which then gets tidied up as

14(x + 4)

by multiplying the 2 and the 7 together. However, since this is a rather long-winded way of doing things, you might as well use the greatest common factor 14 to take out immediately and get straight to the simplest factorisation:

14x + 56 = 14(x + 4).

Introducing Parentheses

Example 1

Observe the following example we saw in the 'Expanding Brackets' tab:

(x + 1)(x + 2) =  + 3x + 2

Let's see how to do this expression in reverse!

+ 3x + 2 = (x + 1)(x + 2)

This is an easy example as the coefficient of x² is 1. In order to do this then, the first thing we do is look at the sign of the constant term, which here is +2. The sign is positive, so now we need to find factors which multiply together to get the constant term, and add together to make the x coefficient - that is, factors which multiply together to make 2 and add together to make 3. There is an obvious answer, which is 1 and 2 (or 2 and 1, it doesn't matter!).

Now, we need to decide if it should be:

• (x + 1) and (x + 2),
• (x - 1) and (x - 2),
• (x - 1) and (x + 2) OR
• (x + 1) and (x - 2).

How do we tell which one to use? We look to the signs of the constant coefficient first - if it is positive, it means that the signs of our 1 and 2 should be the same: either (x + 1), (x + 2) or (x - 1), (x - 2). If the sign was negative, we would need to use differing signs.

Now what? Do we use (x + 1), (x + 2) or (x - 1), (x - 2)? Again, we have a look at the sign of the x coefficient - it is positive, so we will use that sign! Therefore, we are left with (x + 1) and (x + 2). Note that if the x coefficient's sign was negative, we would use the other set instead.

All that's left to do now is to put everything together:

+ 3x + 2 = (x + 1)(x + 2).

Example 2

Let's have a look at another example:

x² - 4x - 12

Again, the coefficient of x² is 1 so this is very straightforward to factorise.

The first thing we do is look to the sign of the constant term: it is negative, so therefore we need two numbers which multiply together to make 12 which have a difference of 4. The answer of course, is 2 and 6. These are the numbers which will lie inside our brackets, so now we need to decide between:

• (x + 2)(x + 6)
• (x - 2)(x - 6)
• (x + 2)(x - 6)
• (x - 2)(x + 6)

Which is the correct answer? Look to the sign of the constant coefficient - it is negative, so we know that our options must have different signs, and our options are now:

• (x + 2)(x - 6)
• (x - 2)(x + 6)

How do we choose which one? Our x coefficient is -4, so now we ask ourselves "What makes -4? Is it 2 - 6 or -2 + 6?" Of course, it is 2 - 6, so we will use the factorisation with +2 and -6, and hence:

x² - 4x - 12 = (x + 2)(x - 6).

Solving For x

When our quadratic expression is equal to some other value, we can use factorising to solve for the unknown variable. This is really easy to do, it only involves one extra step! Have a look at the following expression:

+ x - 12 = 0

Since this expression is equal to something (here, it is equal to 0), we can solve for by first factorising out:

+ x - 12 = (x + 4)(x - 3) = 0

Now we have two expressions, (x + 4) and (x - 3), which multiply together to get 0. Therefore, either:

x + 4 = 0  OR  x - 3 = 0

and hence either:

x = -4  OR  x = 3

Which one is correct? They both are! This is sufficient to answer the question however, as either x = -4 or x = 3 works to solve the quadratic equation. Make sure when answering these types of questions to find all possible values of x, otherwise, your answer is incomplete!

Let's look at some other examples:

x² + 6x + 6 = 0   ⇒   (x + 1)(x + 6) = 0   ⇒   x + 1 = 0 or x + 6 = 0   ⇒   x = -1 or x = -6

- 8x + 16 = 0   ⇒   (x - 4)(x - 4) = 0  ⇒   (x - 4) = 0 or (x - 4) = 0   ⇒   x = 4

x² + x - 2 = 5   ⇒   (x - 1)(x + 2) = 0   ⇒   (x - 1) = 0 or (x + 2) = 0   ⇒   x = 1 or x = -2

When the Coefficient of x² is 0

This is, just like we've seen before, very easy to do! Simply take out and group together 'like' terms to simplify the expression.

Example 1

Observe the following example:

9x + 12.

Each of the coefficients have 3 in common, so we can take this out:

3(3x + 4)

and that's it! That is the most simple form this equation can be written in. Therefore, we have:

9x + 12 = 3(3x + 4).

Example 2

Let's look at another example:

414x - 92

This looks more challenging, however once again there are numbers that these two coefficients have in common: 1, 2, 23, and 46. The greatest common factor is 46, so we will use that one. Therefore, we have:

414x - 92 = 46(9x - 2).

If we are ever not sure of what we have done is right, we can evaluate the right-hand-side of the equation to see if we end up with the left-hand-side...if we can't, something has gone wrong!

46(9x - 2) = (46 x 9)x + (46 x (-2)) = 414x -92

When the Coefficient of x² is Positive (and Not Equal to 1)

While the general form of a quadratic equation is

for unknown constants ab and c and unknown variable x, so far we have only seen cases when a is either 1 or 0. There may be occasions though when we have our a coefficient not equal to 1, making the quadratic equation seem to be more complicated.  There are a few extra things to consider when handling these types of quadratic equations, which will be explored below.

Example 1

Let's have a look at the following quadratic equation:

3x² + 16x + 5

Clearly, we have a = 3, b = 16 and c = 5. We will introduce a new variable p such that

p = a x c

In this example, we have then that

p = 3 x 5 = 15.

Let's list all of the multiplicative pairs of p:

• p = 1 x 15
• p = (-1) x (-15)
• p = 3 x 5
• p = (-3) x (-5)

Since the sign of p is positive, each of the multiplicative pairs have the same sign (either both are positive or both are negative).

Now, have a look at our x coefficient b, which is 16. Both p and b are positive so we are looking for a multiplicative pair which adds up to 16 and have the same sign - this is clearly 1 and 15, so this is the multiplicative pair we will use from now on. Therefore,

16x = x + 15x

Now, we will write our equation as the sum of these two x terms found from this multiplicative pair:

3x² + 16x + 5 = 3x² + x + 15x + 5

We can now group these pairs for factoring (if the grouping doesn't work out first time, swap the two x values around so they do!)

3x² + x + 15x + 5 = (3x² + 15x) + (x + 5)

We can now simplify these:

(3x² + 15x) + (x + 5) = 3x(x + 5) + (x + 5).

This can be simplified even further, as both terms have (x + 5) in common:

3x(x + 5) + (x + 5) = (3x + 1)(x + 5)

That's it! The expression cannot be simplified further, and hence:

3x² + 16x + 5 = (3x + 1)(x + 5)

Is this correct? To check, multiply out the right-hand-side to see if it equals the left-hand-side:

(3x + 1)(x + 5) = 3x² + 15x + x + 5 = 3x² + 16x + 5

Example 2

Let's consider another example:

6x² - 7x - 3

We can see that a = 6, b = 11 and c = 3. We need to create our new variable p such that p = a x c.

p = 6 x (-3) = -18

and then write out the multiplicative pairs:

• p = 1 x (-18)
• p = (-1) x 18
• p = 2 x (-9)
• p = (-2) x 9
• p = 3 x (-6)
• p = (-3) x 6

Since p is negative, each of these multiplicative pairs have differing signs between the pairs. We know that if p was positive, these pairs would have the same signs.

Now, have a look at our x coefficient b, which is 16. Both p and b are negative so we are looking for a multiplicative pair in which the largest number is negative and the smaller is positive (that is, the largest number has the same sign as b) and have different signs, where the numbers have add up to make -7, which is our b value. Considering all of these rules, the pair we want is

• p = 2 x (-9)

To make

-7x = 2x - 9x

Now, we need to write our equation as the sum of these two x terms found from this multiplicative pair:

6x² - 7x - 3 = 6x² + 2x - 9x - 3

We can now group these pairs for factoring. Again, if the grouping doesn't work out first time, swap the two x values around so they do.

6x² - 7x - 3 = (6x² - 9x) + (2x - 3)

6x² - 7x - 3 = 3x(2x - 3) + (2x - 3)

6x² - 7x - 3 = (3x + 1)(2x - 3)

That's it! We can then check that we're right by expanding the right-hand-side and seeing if it is still equal to the left hand side:

(3x + 1)(2x - 3) = 6x² - 9x + 2x -3 = 6x² - 7x - 3

When the Coefficient of x² is Negative

Having a negative a value is exactly the same as having a positive a, except all you need to do now is to accommodate for the negative sign. Following on from above, let's have a look at the following examples.

Example 1

Consider the following equation:

-5x² + 7x - 2

We have a = -5, b = 7 and c = -2. Let's make p such that p = a x c

p = (-5) x (-2) = 10

p is positive, so we can expect its multiplicative pairs to have the same sign.

• p = 1 x 10
• p = (-1) x (-10)
• p = 2 x 5
• p = (-2) x (-5)

Which one is the right one to use? We need to look at our x coefficient b which is 7. Both p and b are positive so we are looking for positive numbers which add up to 7, which of course is the pair:

• p = 2 x 5

Again, we use these to rewrite our x:

7x = 2x + 5x

and hence rewrite our equation:

-5x² + 7x - 2 = -5x² + 2x + 5x - 2

We then group and simplify:

-5x² + 7x - 2 = (-5x² + 5x) + (2x - 2)

-5x² + 7x - 2 = = -5x(x - 1) + 2(x - 1)

-5x² + 7x - 2 = (-5x + 2)(x - 1)

Check:

(-5x + 2)(x - 1) = -5x² + 2x + 5x - 2 = -5x² + 7x - 2

Example 2

Now, let's take things a bit further and consider the following:

"Find all values of x such that 12x² = -18x + 12."

How would we start to solve this problem? We first need to rearrange this equation to make the expression equal 0:

12x² = -18x + 12 ⇒ 0 = -12x² -18x + 12

We could of course have rearranged this to make 12x² +18x - 12 = 0 however we have done the above for the sake of this example showing a negative a value. Indeed, this solving method would have worked with either rearrangement, and we will show that afterwards.

Here, for now, we have a = -12, b = -18 and c = 12.

Let's make p:

p = a x c = (-12) x 12 = -144

p then has the following multiplicative pairs:

• p = 1 x (-144)
• p = (-1) x 144
• p = 2 x (-72)
• p = (-2) x 72
• p = 3 x (-48)
• p = (-3) x 48
• p = 4 x (-36)
• p = (-4) x 36
• p = 6 x (-24)
• p = (-6) x 24
• p = 8 x (-18)
• p = (-8) x 18
• p = 9 x (-16)
• p = (-9) x 16
• p = 12 x (-12)

There are many pairs here! Which one do we use? Once again, we look to the b value, b = -18. Both p and b are negative so we expect our multiplicative pair to have different signs, where the largest number has the same sign as b, and have a difference of 18. The pair which fits all these rules is:

• p = 6 x (-24)

and so we rewrite our x coefficient in terms of this pair:

-18x = 6x - 24x

and so we have:

0 = -12x² -18x + 12 ⇒ 0 = -12x² + 6x - 24x + 12

All we need to do now is regroup and simplify.

0 = -12x² + 6x - 24x + 12 = (-12x² + 6x) + (-24x + 12)

0 = 6x(-2x + 1) + 12(-2x + 1)

0 = (6x + 12)(-2x + 1)

Finally, we solve for x:

0 = 6x + 12 OR 0 = -2x + 1

x = -½ OR x = ½

We can check our answer by inputting our two values of x into our equation to see if it works.

Now, let's consider the alternative arrangement of 12x² = -18x + 12, which is:

12x² +18x - 12 = 0

Let's see if we get the same values of x. Of course, here we have a = 12, b = 18 and c = -12, making:

p = a x c = 12 x (-12) = -144

This is the same p value as last time, and hence we know already that the multiplicative pairs of p are:

• p = 1 x (-144)
• p = (-1) x 144
• p = 2 x (-72)
• p = (-2) x 72
• p = 3 x (-48)
• p = (-3) x 48
• p = 4 x (-36)
• p = (-4) x 36
• p = 6 x (-24)
• p = (-6) x 24
• p = 8 x (-18)
• p = (-8) x 18
• p = 9 x (-16)
• p = (-9) x 16
• p = 12 x (-12)

Our b value is different this time from last time however, as b = 18, not -18. Therefore, we will need to consider a different multiplicative pair...or will we?

Since p is negative we expect our numbers to have different signs. b is positive hence the sign of b goes against the largest number, and we need to consider the numbers which have a difference of b = 18. This is...

• p = (-6) x 24

which is the opposite of what we had earlier! Hence:

18x = -6x + 24x

12x² +18x - 12 =12x² - 6x + 24x - 12 = 0

Once again, we rearrange and simplify:

(12x² - 6x) + (24x - 12) = 0

6x(2x - 1) + 12(2x - 1) = 0

(6x + 12)(2x - 1) = 0

We're not quite done as we still need to solve for x:

6x + 12 = 0 OR 2x - 1 = 0

x = -½ OR x = ½

This is the same as what we had before! Therefore, it doesn't matter which way you rearrange your equation, you will still arrive at your values of x.

ax² + bx

While the general form of a quadratic equation is

for unknown constants ab and c and unknown variable x, there are of course times when we have c equal to zero, and the quadratic equation ends up looking like the form

This is not a problem when this happens as, as it may be easy to see, each term now has an x in common and can be taken out and factorised easily. There may be a common factor that the coefficients of  and x may have as well, which again may be taken out.

x² - 20x = x(x - 20)

14xx² = x(14 - x)

3 - x = x(3x - 1)

14  + 12x = 2x(7x + 6)

-49 - 28x = -7x(7x + 4)

Making Things Even Simpler

Having two x's or unknown variables in an expression can make things hard to work with, so it may be the case that things need to be simplified further in order to work with it easier. However, having a missing c constant can make the factorising difficult. A way we can get around this is by Completing the Square - see that tab on this to make these kinds of equations simpler.

Special Cases When b = 0 and c is a Square

There are special cases when not only b = 0 but also when the c term is a square - this is known as the 'Difference of Two Squares', because we can simplify these expressions into a squared term minus another squared term.

Example 1

For example, take the expression

x² - 9.

This is a simplified way of writing

x² + 0x - 9

such that, where a = 1, b = 0 and c = 9. We can then treat this like an equation we have seen earlier: what two numbers multiply together to make -9 but add up together to make 0? The answer, of course, is x = 3 and x = -3.

Therefore, we have the factorisation to be:

(x + 3)(x - 3).

Example 2

Take the expression:

3x² = 12

Simplifying, this becomes:

3x² - 12 = 0

and simplifying even further we end up with:

3(x² - 4) = 0

We can divide both sides by 3 to obtain

x² - 4 = 0

and then ask ourselves "what two numbers multiply together to make 4 and add together to make 0?" Of course, the answer is 2 and -2, and hence our factorisation will become

(x + 2)(x - 2) = 0

and hence our values of x are

x = -2 and x = 2

Generalisation of the Above

Indeed, for every quadratic equation of the form

x² - d²

for any number d then the factorised form is then

(x - d)(x + d).

So, whenever you have an expression like the above, see if there are any common factors you can factor out.

Sometimes, unfortunately we come across quadratic equations which will not factorise easily. Fortunately, there is still a way we can simplify them! This method is called 'Completing the Square', and involves rewriting the quadratic equation as a square term plus another term, ending up with the form:

Indeed, any quadratic equation can be rewritten this way. How is this in practice? Let's look at a few examples.

Example 1

Let

y = x² + 2x + 7

Let's see if we can rewrite this in the form y = r(x + s)² + t. Note that the rs and t will be different from ab and c!

The first thing we need to do is halve b. b = 2, so half b is 1. This is what will form the inside of our bracket, (x + 1), which will be what we end up squaring.

Now, we know that

(x + 1)² = (x + 1)(x + 1) = x² + 2x + 1

The coefficients of x² and x match our original equation above, but not our constant term. Let's see what this looks like when we negate the constant term:

x² + 2x = x² + 2x + 1 - 1  ⇒  x² + 2x = (x + 1)² - 1

Therefore, when we add our c coefficient, we end up with:

x² + 2x + 7 = (x² + 2x) + 7 = ((x + 1)² - 1) + 7

And hence:

x² + 2x + 7 = (x + 1)² + 6

Therefore, to finish off, we need to remember that the original equation was a function of y, so we shall rewrite y in this format:

y = (x + 1)² + 6

Example 2

Consider the following problem:

"Rewrite x² - 18x + 1 in the form r(x + s)² + t."

We have b = -18 here, so half of b is -9. This is what will be on the inside of our bracket.

(x - 9)

Our next step is to square this and see what we end up with.

(x - 9)² = (x - 9)(x - 9) = x² - 18x + 81

Our coefficients of x² and x here are the same as what we have above, but not the constant term. We will then negate this constant term.

x² - 18x = x² - 18x + 81 - 81  ⇒  x² - 18x = (x - 9)² - 81

And so, when we include our original c coefficient, we end up with:

x² - 18x + 1 = (x² - 18x) + 1 = ((x - 9)² - 81) + 1

Hence:

x² - 18x + 1 = (x - 9)² - 80

c = 0, Continued

When c = 0, completing the square becomes even easier as we can miss off the last few steps completely.

Example 1

Let

x² + 12x

Let's rewrite this in the form r(x + s)² + t.

Here, b = 12 and half of this is 6, so our inside bracket becomes:

(x + 6)

When this is squared, we get:

(x + 6)² = (x + 6)(x + 6) = x² + 12x + 36

This is the same as our original equation except for the constant term 36. So, we will just negate this!

x² + 12x = (x² + 12x + 36) - 36  ⇒  x² + 12x = (x + 6)² - 36

Done!

The formula to solve quadratic equations of the form ax² + bx + c = 0 is given by:

You can use this formula to solve any quadratic equation without having to factorise.

Example 1

"Find the values of x such that 8x² + 2x = 5."

Our first step is to rearrange this equation to make it equal to 0:

8x² + 2x - 5 = 0

and hence our coefficients are a = 8, b = 2 and c = -5. We can simply plug these values into our equation to get the values of x:

The discriminant b² - 4ac > 0 so we can expect this to have two real roots. Evaluating further, we have:

Which then simplifies to

and hence our values of x are

and

Is this right? We can check by inputting these values of x into our original equation and seeing if it makes sense.

Example 2

"Let y = 5x² - 12x - 4. Find all values of x when y = 10."

This question is another way of writing:

10 = 5x² - 12x - 4

which can be rearranged to be:

0 = 5x² - 12x - 14

Therefore, we have a = 5, b = -12 and c = -14. Let's input this into our quadratic formula:

The discriminant b² - 4ac > 0 so we can expect real roots. Evaluating this, we obtain:

And therefore our values of x are:

and