# How do you complete the square 3x^2-6x-1=0?

Feb 4, 2015

$3 \left({x}^{2} - 2 x\right) - 1 = 0 \Rightarrow 3 \left({x}^{2} - 2 x + 1 - 1\right) - 1 = 0 \Rightarrow$
$3 \left({x}^{2} - 2 x + 1\right) - 3 - 1 = 0 \Rightarrow 3 {\left(x - 1\right)}^{2} - 4 = 0 \Rightarrow$
${\left(x - 1\right)}^{2} = \frac{4}{3} \Rightarrow x - 1 = \pm \sqrt{\frac{4}{3}} = 0 \Rightarrow$
$x = 1 \pm \frac{2}{\sqrt{3}} \Rightarrow x = 1 \pm \frac{2 \sqrt{3}}{3} \Rightarrow$
$x = \frac{3 \pm 2 \sqrt{3}}{3}$.