Answer
$\dfrac{-5-i\sqrt{2}}{11}$
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:
$=\dfrac{-15-i\sqrt{18}}{33}
\\=\dfrac{-15-i\sqrt{9(2)}}{33}
\\=\dfrac{-15-i\sqrt{3^2(2)}}{33}
\\=\dfrac{-15-i\cdot 3\sqrt{2}}{33}
\\=\dfrac{-15-3i\sqrt{2}}{33}$
Factor out 3 in the numerator then cancel the common factors to obtain:
$\require{cancel}
\\=\dfrac{3(-5-i\sqrt{2})}{33}
\\=\dfrac{\cancel{3}(-5-i\sqrt{2})}{\cancel{33}11}
\\=\dfrac{-5-i\sqrt{2}}{11}$
