# How do you complete the square to solve 0=9x^2 + 6x - 8?

May 23, 2015

$0 = 9 {x}^{2} + 6 x - 8 = 9 {x}^{2} + 6 x + 1 - 1 - 8$

$= {\left(3 x + 1\right)}^{2} - 9$

Add $9$ to both ends to get:

${\left(3 x + 1\right)}^{2} = 9$

So

$3 x + 1 = \pm \sqrt{9} = \pm 3$

Subtract $1$ from both sides to get:

$3 x = - 1 \pm 3$

Divide both sides by $3$ to get:

$x = - \frac{1}{3} \pm 1$

That is $x = - \frac{4}{3}$ or $x = \frac{2}{3}$

In general $a {x}^{2} + b x + c = a {\left(x + \frac{b}{2 a}\right)}^{2} + \left(c - {b}^{2} / \left(4 a\right)\right)$