Answer
{$-5-i\sqrt {13},-5+i\sqrt {13}$}
Work Step by Step
We know that if $x^{2}=a$, then $x=\pm \sqrt{a}$. Thus, we obtain:
Step 1: $(x+5)^{2}=-13$
Step 2: $x+5=\pm \sqrt {-13}$
Step 3: $x+5=\pm \sqrt {-1\times13}$
Step 4: $x+5=\pm (\sqrt {-1}\times\sqrt {13})$
Step 5: $x+5=\pm (i\times\sqrt {13})$ [as $i=\sqrt {-1}$]
Step 6: $x+5=\pm (i\sqrt {13})$
Step 7: $x=-5\pm (i\sqrt {13})$
Step 8: $x=-5+i\sqrt {13}$ or $x=-5-i\sqrt {13}$
Therefore, the solution set is {$-5-i\sqrt {13},-5+i\sqrt {13}$}.
