Answer
$2+6i\sqrt{7}$
Work Step by Step
RECALL:
(1) $(a-b)^2 = a^2-2ab+b^2$
(2) $\sqrt{-1}=i$
(3) $i^2=-1$
(4) For any real number $a \gt 0$, $\sqrt{-a} = i\sqrt{a}$.
Use rule (4) above to obtain:
$=(-3-i\sqrt{7})^2$
Use rule (1) above with $a=-3$ and $b=i\sqrt{7}$ to obtain:
$=(-3)^2-2(-3)(i\sqrt{7}) + (i\sqrt{7})^2
\\=9+6i\sqrt{7}+i^2(7)
\\=9+6i\sqrt{7} + 7i^2$
Use rule (3) above to obtain:
$=9+6i\sqrt{7} + 7(-1)
\\=9+6i\sqrt{7} -7
\\=2+6i\sqrt{7}$
