Answer
$\left\{ \dfrac{3-\sqrt{29}}{2},\dfrac{3+\sqrt{29}}{2} \right\}$
Work Step by Step
The standard form of the given equation, $
\dfrac{x^2}{3}-x=\dfrac{5}{3}
,$ is
\begin{array}{l}\require{cancel}
3\left( \dfrac{x^2}{3}-x \right)=\left( \dfrac{5}{3} \right)3
\\\\
1(x^2)+3(-x)=5(1)
\\\\
x^2-3x=5
\\\\
x^2-3x-5=0
.\end{array}
Using $\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ (or the Quadratic Formula), the solutions of the quadratic equation, $
x^2-3x-5=0
,$ are
\begin{array}{l}\require{cancel}
\dfrac{-(-3)\pm\sqrt{(-3)^2-4(1)(-5)}}{2(1)}
\\\\=
\dfrac{3\pm\sqrt{9+20}}{2}
\\\\=
\dfrac{3\pm\sqrt{29}}{2}
.\end{array}
Hence, the solutions are $
\left\{ \dfrac{3-\sqrt{29}}{2},\dfrac{3+\sqrt{29}}{2} \right\}
.$
