Answer
{$1 - \dfrac{\sqrt{3}}{3},1 + \dfrac{\sqrt{3}}{3}$}
Work Step by Step
Given: $3x^2-6x+2=0$
This can be re-written as: $x^2-2x=\dfrac{-2}{3}$
We will have to add both sides $(-1)^2$ to complete the square.
Thus, $x^2-2x+(1)^2=\dfrac{-2}{3}+(-1)^2$
or, $(x-1)^2=\dfrac{1}{3}$
or, $(x-1)=\pm \dfrac{\sqrt{3}}{3}$
or, $x=1 \pm \dfrac{\sqrt{3}}{3}$
Hence, our desired solution set is {$1 - \dfrac{\sqrt{3}}{3},1 + \dfrac{\sqrt{3}}{3}$}
