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