Solutions for equations are found when an expression evaluates to zero. One method for solving a quadratic equation is by factorising. After factorising, find the value of `x` that makes each bracket zero.

Rearrange the equation so that one side of the equation is equal to zero. Next, factorise the expression. Then find the *two* values of `x` that will solve the equation. In some instances the two values may be the same - `x^2+6x+9` factorises to `(x+3)(x+3)` which gives you solutions of -3 and -3.

Note that not all quadratic expressions have a solution: for example, `x^2-3x+7` cannot be solved except by using advanced mathematics.

Solve `x^2-4x-5=0`

`x^2-4x-5` = 0

Factorise the expression:

`(x+1)(x-5) = 0`

Determine what values of `x` causes each set of brackets to equal zero:

`(x+1)=0` when `x=-1`, and `(x-5) =0` when `x=5`

The solutions are `x=-1 text( and ) x=5`

Check:

(-1)^{2} - 4(-1) - 5 =0✔

(50^{2} - 4(5) - 5 = 0✔

Answer: `x=-1 text( and ) x=5`

Solve `2x^2-7x=4`

`2x^2-7x=4`

Rearrange the equation so that one side of the equation equals zero:

`2x^2 - 7x - 4 = 0`

Factorise the equation:

`(2x+1)(x-4)=0`

What values of `x` will make each bracket zero?

`(2x+1)=0` when `x=-0.5`

and `(x-4) =0` when `x=4`

The solutions are therefore `x=-0.5 text( and ) x=4`

Check:

2(-0.5)^{2} - 7(-0.5) = 4✔

2(4)^{2} - 7(4) = 4✔

Answer: `x=-frac(1)(2) text( and ) x=4`

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