# How do you solve -2x ^ { 2} = 2x - 4?

Oct 6, 2017

$x = 1 , - 2$

#### Explanation:

rearrange to

$a {x}^{2} + b x + c = 0$

$- 2 {x}^{2} = 2 x - 4$

$\implies 2 {x}^{2} + 2 x - 4 = 0$

divide by 2

${x}^{2} + x - 2 = 0$

factorise

$\left(x + 2\right) \left(x - 1\right) = 0$

giving

$x + 2 = 0 \implies x = - 2$

$x - 1 = 0 \implies x = 1$

Oct 6, 2017

$x = - 2 \mathmr{and} x = 1$

#### Explanation:

To solve this, you first need to put everything on one side:

$- 2 {x}^{2} = 2 x - 4$
$2 {x}^{2} + 2 x - 4 = 0$

From here, we can factor $2$ out:

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

Dividing by two, we get:

${x}^{2} + x - 2 = 0$

This can then be factorised by finding two numbers that multiply to get $- 2$ and add to get $1$. Looking at the factors of $2$, you can find these numbers are $- 1$ and $2$.

$\left(x - 1\right) \left(x + 2\right) = 0$

Now, you have two things multiplied together that equal $0$. This means that one of them has to equal $0$. To find the values of $x$ for this to happen, we split the two parts:

$x - 1 = 0$
$\therefore x = 1$

OR

$x + 2 = 0$
$\therefore x = - 2$

So, the solution to this equation is $x = - 2 \mathmr{and} x = 1$