Answer
$\left\{ \dfrac{3-\sqrt{11}}{12},\dfrac{3+\sqrt{11}}{12} \right\}$
Work Step by Step
The standard form of the given equation, $
\dfrac{1}{3}y^2=y+\dfrac{1}{6}
,$ is
\begin{array}{l}\require{cancel}
6\left( \dfrac{1}{3}y^2 \right)=\left( y+\dfrac{1}{6} \right)6
\\\\
2(y^2)=y(6)+1(1)
\\\\
2y^2=6y+1
\\\\
2y^2-6y-1=0
.\end{array}
Using $\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ (or the Quadratic Formula), the solutions of the quadratic equation, $
2y^2-6y-1=0
,$ are
\begin{array}{l}\require{cancel}
\dfrac{-(-6)\pm\sqrt{(-6)^2-4(2)(-1)}}{2(2)}
\\\\=
\dfrac{-(-6)\pm\sqrt{(-6)^2-4(2)(-1)}}{4}
\\\\=
\dfrac{6\pm\sqrt{36+8}}{24}
\\\\=
\dfrac{6\pm\sqrt{44}}{24}
\\\\=
\dfrac{6\pm\sqrt{4\cdot11}}{24}
\\\\=
\dfrac{6\pm\sqrt{(2)^2\cdot11}}{24}
\\\\=
\dfrac{6\pm2\sqrt{11}}{24}
\\\\=
\dfrac{2(3\pm\sqrt{11})}{24}
\\\\=
\dfrac{\cancel{2}(3\pm\sqrt{11})}{\cancel{2}\cdot12}
\\\\=
\dfrac{3\pm\sqrt{11}}{12}
.\end{array}
Hence, the solutions are $
\left\{ \dfrac{3-\sqrt{11}}{12},\dfrac{3+\sqrt{11}}{12} \right\}
.$