Answer
$\left\{ 3-\sqrt{6},3+\sqrt{6} \right\}$
Work Step by Step
The standard form of the given equation, $
\dfrac{1}{6}x^2+x+\dfrac{1}{3}=0
,$ is
\begin{array}{l}\require{cancel}
6\left( \dfrac{1}{6}x^2+x+\dfrac{1}{3} \right)=(0)6
\\\\
1(x^2)+6(x)+2(1)=0
\\\\
x^2+6x+2=0
.\end{array}
Using $\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ (or the Quadratic Formula), the solutions of the quadratic equation, $
x^2+6x+2=0
,$ are
\begin{array}{l}\require{cancel}
\dfrac{-(-6)\pm\sqrt{(-6)^2-4(1)(2)}}{2(1)}
\\\\=
\dfrac{6\pm\sqrt{36-8}}{2}
\\\\=
\dfrac{6\pm\sqrt{24}}{2}
\\\\=
\dfrac{6\pm\sqrt{4\cdot6}}{2}
\\\\=
\dfrac{6\pm\sqrt{(2)^2\cdot6}}{2}
\\\\=
\dfrac{6\pm2\sqrt{6}}{2}
\\\\=
\dfrac{2(3\pm\sqrt{6})}{2}
\\\\=
\dfrac{\cancel{2}(3\pm\sqrt{6})}{\cancel{2}}
\\\\=
3\pm\sqrt{6}
.\end{array}
Hence, the solutions are $
\left\{ 3-\sqrt{6},3+\sqrt{6} \right\}
.$
