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