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

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Answer 1 · 2015-07-03T12:27:04

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

$->x^2+4x-24=0$

Since $(x+a)^2=x^2 +2ax+a^2$
We take half of the coefficient of $x$ , and square it:
$x^2+4x+4=(x+2)^2$

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

So $x+2=sqrt28=2sqrt7->x=-2+2sqrt7$
Or $x+2=-sqrt28=-2sqrt7->x=-2-2sqrt7$

(sometimes written as $x_(1,2)=-2+-2sqrt7$ )

How do you solve 5x^2 + 20x - 120 = 05x^2 + 20x - 120 = 0 using completing the square?