# How do you solve 0=x^2-3x-6 by completing the square?

##### 1 Answer
May 17, 2015

$0 = {x}^{2} - 3 x - 6 = \left({x}^{2} - 3 x + \frac{9}{4}\right) - \frac{9}{4} - 6$

$= {\left(x - \frac{3}{2}\right)}^{2} - \frac{33}{4}$

In general:

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

Adding $\frac{33}{4}$ to both sides we get

${\left(x - \frac{3}{2}\right)}^{2} = \frac{33}{4}$

So $x - \frac{3}{2} = \pm \sqrt{\frac{33}{4}} = \pm \frac{\sqrt{33}}{2}$

Adding $\frac{3}{2}$ to both sides we get

$x = \frac{3}{2} \pm \frac{\sqrt{33}}{2} = \frac{3 \pm \sqrt{33}}{2}$