# How to Solve a Quadratic Equation - Three Methods for Solving Quadratic Equations

*I am a former maths teacher and owner of DoingMaths. I love writing about maths, its applications, and fun mathematical facts.*

## What Is a Quadratic Equation?

A quadratic equation is any equation that can be rearranged into the form ax^{2}+bx+c=0 where a, b and c are numbers with a ≠ 0 (if a = 0 we get a linear equation bx+c=0).

To solve a quadratic we are looking for the values of x which make ax^{2}+bx+c=0 true. There will always be either two real solutions, two complex number solutions or exactly one solution (this is known as a double root).

In this article we will look at the three most popular algebraic methods for finding these solutions.

## Method 1 - Factorising the Equation

If the quadratic equation has real, rational solutions, the quickest way to solve it is often to factorise into the form (px + q)(mx + n), where m, n, p and q are integers. This is especially true where the coefficient of x^{2} is 1.

__Example 1__ - Solve x^{2}+7x+12=0

To factorise, we are looking for two numbers that multiply to make 12 and add to make 7. This is 3 and 4, so we get x^{2}+7x+12=(x+3)(x+4) = 0.

We now have two brackets multiplying to make 0, hence our solutions must be when these brackets each equal 0. Our solutions are therefore x=−3 and x=−4.

__Example 2__ - Solve 6x^{2}+11x−10=0

A slightly trickier one to factorise, but by first considering the 6 and −10 and with some trial and error, we can see that 6x^{2}+11x−10=(2x+5)(3x−2)=0. Again we can see that the brackets must equal zero and so x=−5/2 and x=2/3.

__Example 3__ - Solve x^{2}−36=0

An expansion of this method is when we have the difference of two squares. In this example we have the square of x minus the square of 6. It can be seen quite quickly that x^{2}−36 = (x+6)(x−6) and therefore we get the answers −6 and 6.

## Method 2 - Completing the Square

A similar method to above, completing the square also involves factorising, but this method will also work for complex and/or irrational answers. This time we are looking to convert our quadratic into the form (x+q)^{2}+r=0 where q and r are real numbers to be found. Let's start with the same example as above.

__Example 1__ - Solve x^{2}+7x+12=0

Consider (px+q)^{2}. When we expand this we get p^{2}x^{2}+2pqx+q^{2}. In our example, the coefficient of x^{2} is 1, so p=1. The coefficient of x is 7, so 2pq=7 and hence q = 7/2.

Our expression must therefore begin (x+7/2)^{2}. If we expand this we get (x+7/2)^{2 }= x^{2}+7x+49/4 and so, by rearranging we can see that x^{2}+7x = (x+7/2)^{2} − 49/4.

Therefore x^{2}+7x−12 = (x+7/2)^{2} − 49/4 + 12

=(x+7/2)^{2} − 1/4.

Setting this equal to 0 and rearranging, we get:

(x+7/2)^{2} − 1/4 = 0

(x+7/2)^{2} = 1/4

x+7/2 = ±1/2

x = −7/2 ±1/2

x = −3 and −4

__Example 2__ - Solve 2x^{2}−3x−4=0

This time we have a coefficient of x^{2} which is not 1. In order to complete the square on these types of quadratics, it is easiest to first divide by this coefficient. In this case, dividing by 2 gives us:

x^{2}−(3/2)x−2=0

By our method from above, we put half of 3/2 into the bracket and then remove the square of this to cancel out the extra part given when the bracket is expanded, so:

x^{2}−(3/2)x−2=(x−3/4)^{2}−(3/4)^{2}−2

=(x−3/4)^{2}−41/16

Setting this equal to 0 and rearranging gives us:

(x−3/4)^{2}−41/16=0

(x−3/4)^{2}=41/16

x−3/4=±(√41)/4

x=3/4 ±(√41)/4 (approximately −0.851 and 2.351)

## Method 3 - The Quadratic Formula

Our third method works for all quadratic equations whether their solutions are rational or irrational, real or complex. The quadratic formula is a set formula which can be derived by completing the square on the quadratic ax^{2}+bx+c=0. To see how this is done, watch the video below.

## How to Use the Quadratic Equation

For any quadratic equation of the form ax^{2}+bx+c=0, the solutions can be found by using the formula x = (−b ± √(b^{2} − 4ac))/2a.

__Example 1__ - Solve x^{2}+7x+12=0

Using the quadratic formula with a = 1, b = 7 and c = 12:

x = (−7 ± √(7^{2} − 4 × 1 × 12)) / (2 × 1)

= (−7 ± √(49 − 48)) / 2

= (−7 ± √1) / 2

= (−7 ± 1) / 2

= −4 or −3

__Example 2__ - Solve 2x^{2}−3x−4=0

x = (3 ± √((−3)^{2} − 4 × 2 × −4)) / (2 × 2)

= (3 ± √(9 + 32)) / 4

= (3 ± √41) / 4

= 3/4 + (√41)/4

*This content is accurate and true to the best of the author’s knowledge and is not meant to substitute for formal and individualized advice from a qualified professional.*

**© 2021 David**

## Comments

**David (author)** from West Midlands, England on June 08, 2021:

Thank you.

**Nayanjyoti Mirdha** on June 08, 2021:

Interesting content.Keep writing!!!