Answer
$5 \text{ (double root)}$
Work Step by Step
The standard form of the given equation, $
\dfrac{2}{3}x^2-\dfrac{20}{3}x=-\dfrac{100}{6}
,$ is
\begin{array}{l}\require{cancel}
6\left( \dfrac{2}{3}x^2-\dfrac{20}{3}x \right)=\left( -\dfrac{100}{6} \right)6
\\\\
2(2x^2)-2(20x)=-100(1)
\\\\
4x^2-40x=-100
\\\\
4x^2-40x+100=0
\\\\
\dfrac{4x^2-40x+100}{4}=\dfrac{0}{4}
\\\\
x^2-10x+25=0
.\end{array}
Using $\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ (or the Quadratic Formula), the solutions of the quadratic equation, $
x^2-10x+25=0
,$ are
\begin{array}{l}\require{cancel}
\dfrac{-(-10)\pm\sqrt{(-10)^2-4(1)(25)}}{2(1)}
\\\\=
\dfrac{10\pm\sqrt{100-100}}{2}
\\\\=
\dfrac{10\pm\sqrt{0}}{2}
\\\\=
\dfrac{10\pm0}{2}
\\\\=
\dfrac{10}{2}
\\\\=
5
.\end{array}
Hence, the solutions are $
5 \text{ (double root)}
.$
