Answer
The solutions are $x=\dfrac{1}{3}\pm\dfrac{\sqrt{7}}{3}$
Work Step by Step
$3x^{2}-2x-2=0$
Take out common factor $3$ from the left side:
$3\Big(x^{2}-\dfrac{2}{3}x-\dfrac{2}{3}\Big)=0$
Take the $3$ to divide the right side:
$x^{2}-\dfrac{2}{3}x-\dfrac{2}{3}=\dfrac{0}{3}$
$x^{2}-\dfrac{2}{3}x-\dfrac{2}{3}=0$
Take $\dfrac{2}{3}$ to the right side:
$x^{2}-\dfrac{2}{3}x=\dfrac{2}{3}$
Add $\Big(\dfrac{b}{2}\Big)^{2}$ to both sides of the equation. In this case, $b=-\dfrac{2}{3}$:
$x^{2}-\dfrac{2}{3}x+\Big(-\dfrac{2}{2\cdot3}\Big)^{2}=\dfrac{2}{3}+\Big(-\dfrac{2}{2\cdot3}\Big)^{2}$
$x^{2}-\dfrac{2}{3}x+\dfrac{1}{9}=\dfrac{2}{3}+\dfrac{1}{9}$
$x^{2}-\dfrac{2}{3}x+\dfrac{1}{9}=\dfrac{7}{9}$
Factor the left side of the equation, which is a perfect square trinomial:
$\Big(x-\dfrac{1}{3}\Big)^{2}=\dfrac{7}{9}$
Take the square root of both sides:
$\sqrt{\Big(x-\dfrac{1}{3}\Big)^{2}}=\pm\sqrt{\dfrac{7}{9}}$
$x-\dfrac{1}{3}=\pm\dfrac{\sqrt{7}}{3}$
Solve for $x$:
$x=\dfrac{1}{3}\pm\dfrac{\sqrt{7}}{3}$
