Answer
$\left\{ -4-2i,-4+2i \right\}$
Work Step by Step
The standard form of the given equation, $
\dfrac{1}{8}x^2+x+\dfrac{5}{2}=0
,$ is
\begin{array}{l}\require{cancel}
8\left( \dfrac{1}{8}x^2+x+\dfrac{5}{2} \right)=(0)8
\\\\
1(x^2)+8(x)+4(5)=0
\\\\
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{-16}}{2}
\\\\=
\dfrac{-8\pm i\sqrt{(4)^2}}{2}
\\\\=
\dfrac{-8\pm 4i}{2}
\\\\=
\dfrac{2(-4\pm 2i)}{2}
\\\\=
\dfrac{\cancel{2}(-4\pm 2i)}{\cancel{2}}
\\\\=
-4\pm 2i
.\end{array}
Hence, the solutions are $
\left\{ -4-2i,-4+2i \right\}
.$
