Answer
$\dfrac{-6+i\sqrt{7}}{16}$
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:
$=\dfrac{-12+i\sqrt{28}}{32}
\\=\dfrac{-12+i\sqrt{4(7)}}{32}
\\=\dfrac{-12+i\sqrt{2^2(7)}}{32}
\\=\dfrac{-12+i\cdot 2\sqrt{7}}{32}
\\=\dfrac{-12+2i\sqrt{7}}{32}$
Factor out 2 in the numerator then cancel the common factors to obtain:
$\require{cancel}
\\=\dfrac{2(-6+i\sqrt{7})}{32}
\\=\dfrac{\cancel{2}(-6+i\sqrt{7})}{\cancel{32}16}
\\=\dfrac{-6+i\sqrt{7}}{16}$
