Answer
$\left\{ -10,2 \right\}$
Work Step by Step
The standard form of the given equation, $
\dfrac{1}{8}x^2+x=\dfrac{5}{2}
,$ is
\begin{array}{l}\require{cancel}
8\left( \dfrac{1}{8}x^2+x \right)=\left( \dfrac{5}{2} \right)8
\\\\
1(x^2)+8(x)=5(4)
\\\\
x^2+8x=20
\\\\
x^2+8x-20=0
.\end{array}
Using $\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ (or the Quadratic Formula), the solutions of the quadratic equation, $
x^2+8x-20=0
,$ are
\begin{array}{l}\require{cancel}
\dfrac{-(8)\pm\sqrt{(8)^2-4(1)(-20)}}{2(1)}
\\\\=
\dfrac{-8\pm\sqrt{64+80}}{2}
\\\\=
\dfrac{-8\pm\sqrt{144}}{2}
\\\\=
\dfrac{-8\pm\sqrt{(12)^2}}{2}
\\\\=
\dfrac{-8\pm12}{2}
\\\\=
\dfrac{-8-12}{2}
\text{ OR }
\dfrac{-8+12}{2}
\\\\=
\dfrac{-20}{2}
\text{ OR }
\dfrac{4}{2}
\\\\=
-10
\text{ OR }
2
.\end{array}
Hence, the solutions are $
\left\{ -10,2 \right\}
.$
