Answer
$\dfrac{-3+i\sqrt{3}}{24}$
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:
$=\dfrac{-6+i\sqrt{12}}{48}
\\=\dfrac{-6+i\sqrt{4(3)}}{48}
\\=\dfrac{-6+i\sqrt{2^2(3)}}{48}
\\=\dfrac{-6+i\cdot 2\sqrt{3}}{48}
\\=\dfrac{-6+2i\sqrt{3}}{48}$
Factor out 2 in the numerator then cancel the common factors to obtain:
$\require{cancel}
\\=\dfrac{2(-3+i\sqrt{3})}{48}
\\=\dfrac{\cancel{2}(-3+i\sqrt{3})}{\cancel{48}24}
\\=\dfrac{-3+i\sqrt{3}}{24}$
