Answer
{$-4-2i\sqrt {7},-4+2i\sqrt {7}$}
Work Step by Step
We know that if $x^{2}=a$, then $x=\pm \sqrt{a}$. Thus, we obtain:
Step 1: $(x+4)^{2}=-28$
Step 2: $x+4=\pm \sqrt {-28}$
Step 3: $x+4=\pm \sqrt {-1\times4\times7}$
Step 4: $x+4=\pm (\sqrt {-1}\times\sqrt {4}\times\sqrt 7)$
Step 5: $x+4=\pm (i\times2\times\sqrt {7})$ [as $i=\sqrt {-1}$]
Step 6: $x+4=\pm (2i\sqrt {7})$
Step 7: $x=-4\pm (2i\sqrt {7})$
Step 8: $x=-4+2i\sqrt {7}$ or $x=-4-2i\sqrt {7}$
Therefore, the solution set is {$-4-2i\sqrt {7},-4+2i\sqrt {7}$}.
