# How do you solve 2x^2-2x-12=0?

Aug 15, 2016

$x = 3 \mathmr{and} x = - 2$

#### Explanation:

All the numbers in the equation are even, so divide through by 2.

${x}^{2} - x - 6 = 0 \text{ find factors}$

The factors of 6 which subtract to make 1 are 2 and 2.

$\left(x - 3\right) \left(x + 2\right) = 0$

If the answer to multiplication is 0, then one of the factors must be 0,

If $x - 3 = 0 , \text{then } x = 3$

If $x + 2 = 0 , \text{then } x = - 2$

Aug 15, 2016

$x = - 2 , x = 3$

#### Explanation:

The first step is to take out a $\textcolor{b l u e}{\text{common factor}}$ of 2.

$\Rightarrow 2 \left({x}^{2} - x - 6\right) = 0$

We now require to factorise the $\textcolor{b l u e}{\text{quadratic}}$ in the
bracket.

Find the factors of - 6 which sum to - 1.

These are - 3 and +2

$\Rightarrow 2 {x}^{2} - 2 x - 12 = 2 \left(x - 3\right) \left(x + 2\right) = 0$

Either of the factors (x - 3) , (x +2) can equal zero.

solve : $x - 3 = 0 \Rightarrow x = 3$

solve : $x + 2 = 0 \Rightarrow x = - 2$