# How do you solve 5x^2 + 20x - 120 = 0 using completing the square?

Jul 3, 2015

Logical first step would be to divide everything by $5$

#### Explanation:

$\to {x}^{2} + 4 x - 24 = 0$

Since ${\left(x + a\right)}^{2} = {x}^{2} + 2 a x + {a}^{2}$
We take half of the coefficient of $x$, and square it:
${x}^{2} + 4 x + 4 = {\left(x + 2\right)}^{2}$

We have to balance the $4$ we used with the $- 24$
So the equation becomes:
${x}^{2} + 4 x + 4 - 28 = 0 \to$ add 28:
${\left(x + 2\right)}^{2} = 28$

So $x + 2 = \sqrt{28} = 2 \sqrt{7} \to x = - 2 + 2 \sqrt{7}$
Or $x + 2 = - \sqrt{28} = - 2 \sqrt{7} \to x = - 2 - 2 \sqrt{7}$

(sometimes written as ${x}_{1 , 2} = - 2 \pm 2 \sqrt{7}$)