Answer
$\dfrac{-6i}{3+2i}=-\dfrac{12}{13}-\dfrac{18}{13}i$
Work Step by Step
$\dfrac{-6i}{3+2i}$
Multiply the numerator and the denominator of the fraction by the complex conjugate of the denominator:
$\dfrac{-6i}{3+2i}=\dfrac{-6i}{3+2i}\cdot\dfrac{3-2i}{3-2i}=\dfrac{-6i(3-2i)}{3^{2}-(2i)^{2}}=\dfrac{-18i+12i^{2}}{9-4i^{2}}=...$
Substitute $i^{2}$ by $-1$ and simplify:
$...=\dfrac{-18i+12(-1)}{9-4(-1)}=\dfrac{-18i-12}{9+4}=\dfrac{-12-18i}{13}=...$
$...=-\dfrac{12}{13}-\dfrac{18}{13}i$
