Answer
$x=\left\{ \dfrac{1-2\sqrt{3}}{3},\dfrac{1+2\sqrt{3}}{3} \right\}
$
Work Step by Step
Taking the square root of both sides, the solutions to the given equation, $
(3x-1)^2=12
,$ are
\begin{array}{l}\require{cancel}
3x-1=\pm\sqrt{12}
\\\\
3x-1=\pm\sqrt{4\cdot3}
\\\\
3x-1=\pm\sqrt{(2)^2\cdot3}
\\\\
3x-1=\pm2\sqrt{3}
\\\\
3x=1\pm2\sqrt{3}
\\\\
3x=1\pm2\sqrt{3}
\\\\
x=\dfrac{1\pm2\sqrt{3}}{3}
.\end{array}
Hence, the solutions are $
x=\left\{ \dfrac{1-2\sqrt{3}}{3},\dfrac{1+2\sqrt{3}}{3} \right\}
.$