Answer
$-\dfrac{12}{13} - \dfrac{18}{13}i$
Work Step by Step
Rationalize the denominator by multiplying the conjugate of the denominator, which is $3-2i$, to both the numerator and the denominator to obtain:
$=\dfrac{-6i(3-2i)}{(3+2i)(3-2i)}
\\=\dfrac{-18i+12i^2}{(3+2i)(3-2i)}$
Use the rule $(a-b)(a+b) = a^2-b^2$ to obtain:
$=\dfrac{-18i+12i^2}{3^2-(2i)^2}
\\=\dfrac{-18i+12i^2}{9-4i^2}$
Use the fact that $i^2=-1$ to obtain:
$=\dfrac{-18i+12(-1)}{9-4(-1)}
\\=\dfrac{-18i-12}{9+4}
\\=\dfrac{-12-18i}{13}
\\=-\dfrac{12}{13} - \dfrac{18}{13}i$
