How do you solve 3x^2-6x-1=0 using completing the square?

Jul 7, 2015

$y = 3 {x}^{2} - 6 x - 1 = 0$

Explanation:

$y = 3 \left({x}^{2} - 2 x - \frac{1}{3}\right) = 0$

$\left({x}^{2} - 2 x\right) = \frac{1}{3}$
$\left({x}^{2} - 2 x + 1\right) = \frac{1}{3} + 1$
${\left(x - 1\right)}^{2} = \frac{4}{3}$
$\left(x - 1\right) = \pm \frac{2}{\sqrt{3}} = \frac{2 \sqrt{3}}{3}$

$x = 1 \pm \frac{2 \sqrt{3}}{3}$