Answer
$x=1\pm\dfrac{2\sqrt{3}}{3}$
Work Step by Step
$3x^{2}-6x-1=0$
Take the independent term to the right side of the equation:
$3x^{2}-6x=1$
Take out common factor $3$ on the left side of the equation:
$3(x^{2}-2x)=1$
Let's complete the square of the expression inside the parentheses. Remember that, in order to complete the square, we need to add $(\dfrac{b}{2})^{2}$ to both sides of the equation. $b$ is always the coefficient of the first degree term. For the expression inside the parentheses, $b=-2$.
Please note that, since the expression whose square needs to be completed is multiplied by $3$, we will add $3(\dfrac{b}{2})^{2}$ to right side of the equation.
$3[x^{2}-2x+(\dfrac{-2}{2})^{2}]=1+3(\dfrac{-2}{2})^{2}$
$3(x^{2}-2x+1)=1+3$
$3(x^{2}-2x+1)=4$
$x^{2}-2x+1=\dfrac{4}{3}$
We have a perfect square trinomial on the left side of the equation. We factor it and the equation becomes:
$(x-1)^{2}=\dfrac{4}{3}$
Take the square root of both sides of the equation:
$\sqrt{(x-1)^{2}}=\sqrt{\dfrac{4}{3}}$
$x-1=\pm\dfrac{2}{\sqrt{3}}$
Rationalizing the right side of the equation, and solving for $x$ we get:
$x=1\pm\dfrac{2\sqrt{3}}{3}$
