# How do you solve -20x^2 - 20x + 40 = 0?

Jul 3, 2015

You may start by dividing everything by $20$
It also helps if you reverse all the signs.

#### Explanation:

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

This can be factorised:
$\to \left(x - 1\right) \cdot \left(x + 2\right) = 0$
$\to x = 1 \mathmr{and} x = - 2$

Jul 3, 2015

I found:
${x}_{1} \equiv 1$
${x}_{2} = - 2$

#### Explanation:

Again, you can use the Quadratic Formula but we can also try something...interesting!
Divide all by $20$;
$- {x}^{2} - x + 2 = 0$
Rearrange:
$- {x}^{2} - x = - 2$
again...
${x}^{2} + x = 2$
add and subtract $\frac{1}{4}$;
${x}^{2} + x + \frac{1}{4} - \frac{1}{4} = 2$
rearrange:
${x}^{2} + x + \frac{1}{4} = 2 + \frac{1}{4}$
${\left(x + \frac{1}{2}\right)}^{2} = \frac{9}{4}$
root square both sides:
$x + \frac{1}{2} = \pm \sqrt{\frac{9}{4}} = \pm \frac{3}{2}$
so you get two solutions:
${x}_{1} = - \frac{1}{2} + \frac{3}{2} = 1$
${x}_{2} = - \frac{1}{2} - \frac{3}{2} = - 2$