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

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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 ax2+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 ax2+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 x2 is 1.

### Example 1 - Solve x2+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 x2+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 6x2+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 6x2+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 x2−36=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 x2−36 = (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 x2+7x+12=0

Consider (px+q)2. When we expand this we get p2x2+2pqx+q2. In our example, the coefficient of x2 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 = x2+7x+49/4 and so, by rearranging we can see that x2+7x = (x+7/2)2 − 49/4.

Therefore x2+7x−12 = (x+7/2)2 − 49/4 + 12

=(x+7/2)2 − 1/4.

Setting this equal to 0 and rearranging, we get:

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(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 2x2−3x−4=0

This time we have a coefficient of x2 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:

x2−(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:

x2−(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 ax2+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 ax2+bx+c=0, the solutions can be found by using the formula x = (−b ± √(b2 − 4ac))/2a.

### Example 1 - Solve x2+7x+12=0

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

x = (−7 ± √(72 − 4 × 1 × 12)) / (2 × 1)

= (−7 ± √(49 − 48)) / 2

= (−7 ± √1) / 2

= (−7 ± 1) / 2

= −4 or −3

### Example 2 - Solve 2x2−3x−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.