# How do you solve using the completing the square method 9x^2-12x-2=0?

Jul 29, 2016

$x = \frac{2 + \sqrt{6}}{3}$
$x = \frac{2 - \sqrt{6}}{3}$

#### Explanation:

$9 {x}^{2} - 12 x - 2 = 0$
or
${\left(3 x\right)}^{2} - 2 \left(3 x\right) \left(2\right) + {2}^{2} - {2}^{2} - 2 = 0$
or
${\left(3 x - 2\right)}^{2} = 6$
or
$3 x - 2 = \pm \sqrt{6}$
or
$3 x = 2 \pm \sqrt{6}$
or
$x = \frac{2 \pm \sqrt{6}}{3}$
or
$x = \frac{2 + \sqrt{6}}{3}$
or
$x = \frac{2 - \sqrt{6}}{3}$