Answer
{$2-4i,2+4i$}
Work Step by Step
We know that if $x^{2}=a$, then $x=\pm \sqrt{a}$. Thus, we obtain:
Step 1: $(x-2)^{2}=-16$
Step 2: $x-2=\pm \sqrt {-16}$
Step 3: $x-2=\pm \sqrt {-1\times16}$
Step 4: $x-2=\pm (\sqrt {-1}\times\sqrt {16})$
Step 5: $x-2=\pm (i\times4)$ [as $i=\sqrt{-1}$]
Step 6: $x-2=\pm 4i$
Step 7: $x-2=4i$ or $x-2=-4i$
Step 8: $x=2+4i$ or $x=2-4i$
The solution set is {$2-4i,2+4i$}.
