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