Answer
$\left\{ -1-i\sqrt{3},-1+i\sqrt{3} \right\}$
Work Step by Step
The standard form of the given equation, $
(n-2)^2=2n
,$ is
\begin{array}{l}\require{cancel}
(n)^2-2(n)(-2)+(-2)^2=2n
\\\\
n^2+4n+4=2n
\\\\
n^2+4n-2n+4=0
\\\\
n^2+2n+4=0
.\end{array}
Using $\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ (or the Quadratic Formula), the solutions of the quadratic equation, $
n^2+2n+4=0
,$ are
\begin{array}{l}\require{cancel}
\dfrac{-(2)\pm\sqrt{(2)^2-4(1)(4)}}{2(1)}
\\\\=
\dfrac{-2\pm\sqrt{4-16}}{2}
\\\\=
\dfrac{-2\pm\sqrt{-12}}{2}
\\\\=
\dfrac{-2\pm\sqrt{-1}\sqrt{12}}{2}
\\\\=
\dfrac{-2\pm i\sqrt{4\cdot3}}{2}
\\\\=
\dfrac{-2\pm i\sqrt{(2)^2\cdot3}}{2}
\\\\=
\dfrac{-2\pm 2i\sqrt{3}}{2}
\\\\=
\dfrac{2(-1\pm i\sqrt{3})}{2}
\\\\=
\dfrac{\cancel{2}(-1\pm i\sqrt{3})}{\cancel{2}}
\\\\=
-1\pm i\sqrt{3}
.\end{array}
Hence, the solutions are $
\left\{ -1-i\sqrt{3},-1+i\sqrt{3} \right\}
.$
