There are three ways of solving quadratic equations: Factorisation, Completing the Square and using the Quadratic Equation. All these methods are correct and it is up to you to choose which one you prefer to use for each type of each exercise!

To easily reference the coefficients we will use the standard form of the quadratic equation: ax^{2} + bx + c = 0. In our example a = 1, b = -10 and c = 21. Watchout for the minus sign in b!

**Factorisation**

Factorisation consists in splitting the quadratic equation into two brackets. This method is most commonly used when the coefficient of x^{2 }(**a** in the standard form) equals 1. To factorise the equation we need to find a set of two numbers that when added equal **b** (-10 in this case) and when multiplied equal **c** (21 in this case). The solutions are -7 and -3 since (-7) + (-3) = -10 and (-7)*(-3) = 21. As always, it's very important to look out for the minus signs! To finalise include the obtained numbers in the two set of brackets, this will result in the following solution:

(x - 7)*(x - 3) = 0

You can check this is correct by multiplying the two brackets and checking you are obtaining the original quadratic equation.

Now you only need to check which values of x would make x equal 0 in each bracket, in this case **x= 7 and x=3.**

**Completing the Square**

Completing the Square consists on leaving the quadratic equation in the form of a squared bracket and a number, like for an example in the following equation: ( x + m)^{2} + n = 0 (Note **m** and **n** will be numbers)

To do this, we need to firstly analyse the part of the equation that has an x, in this case x^{2}- 10x , ignore the +21 for now. In particularl we want to look at the coefficient **b**=-10, divide it by 2 and square the result:

-10 / 2 = -5 Note this will represent the** m **from our example

(-5)^{2 }= 25 Note that a squared minus sign equals - * - = +

We now need to include this number in our equation, but we can't just insert a random number and thus we will need to also include a number with the oposite sign ( as this would equal 0 if solved). We then take the part of the equation that includes the **m and 25** and turn it into a squared bracket. This will result in:

x^{2}- 10x = **x**^{2}- 10x + 25 - 25 = **( x - 5)**^{2} -25 Note that the bold parts are equal

To finish this method, we need to substitute the previous result in the original equation, do not forget the +21! Use the equation derived above to complete the quadratic equaiton.

**x**^{2}- 10x + 21 = **( x - 5)**^{2} - 25 + 21 = 0 Note that the bold parts are equal

( x - 5)^{2} - 25 + 21 = 0

( x - 5)^{2} - 4 = 0

We can now solve the equation by rearranging the numbers. Firstly, let's isolate **n**= -4

( x - 5)^{2} = 4 The inverse of a degree of 2 is a square root, and thus:

( x - 5) = √4 = ± 2 Note the root of 4 can be either 2 or -2 as (2*2) = (-2 * -2) = 4

x = 5 ± 2 This equation has two solutions, x = 5 + 2 = **7** and x = 5 - 2 = **3.**

__Quadratic Formula__

The quadratic formula is defined as:

x =( -b ± √(b^{2 }- 4ac) ) / 2a

This can seem complicated but after using it a few times you will get the hang of it and you'll be able to solve any quadratic equation!

To use it, simply substitute a, b and c by all the coefficients in the equation, this will result in:

x = ( -(-10) ± √((-10)^{2} - 4*1*21) ) / 2*1

= ( 10 ± √ 100 - 84 ) / 2 = ( 10 ± √ 16 ) 2 = (10 ± √ 16 ) / 2

= (10 ± 4 ) / 2

Solving this will give us the following two solutions:

x = (10 + 4) /2 = 14/2 = **7**

x = (10 - 4) /2 = 6/2 = **3**

**Conclusion**

As demonstrated above, all three methods can be used to obtain the same results , x=7 and x=3. It is therefore up to you to try them and see which one works best for you! Quadratic equations appear in a wide range of GCSE exercises and the best way to avoid small mistakes is by practicing different exercises.