#### Answer

$-1+2i$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To divide the given expression, $
\dfrac{-7+4i}{3+2i}
,$ multiply both the numerator and the denominator by the complex conjugate of the denominator. Use special products to multiply the resulting expression and use $i^2=-1.$
$\bf{\text{Solution Details:}}$
Multiplying both the numerator and the denominator by the complex conjugate of the denominator, the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{-7+4i}{3+2i}\cdot\dfrac{3-2i}{3-2i}
\\\\=
\dfrac{(-7+4i)(3-2i)}{(3+2i)(3-2i)}
.\end{array}
Using the product of the sum and difference of like terms which is given by $(a+b)(a-b)=a^2-b^2,$ the expression above is equivalent
\begin{array}{l}\require{cancel}
\dfrac{(-7+4i)(3-2i)}{(3)^2-(2i)^2}
\\\\=
\dfrac{(-7+4i)(3-2i)}{9-4i^2}
.\end{array}
Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to\begin{array}{l}\require{cancel}
\dfrac{-7(3)-7(-2i)+4i(3)+4i(-2i)}{9-4i^2}
\\\\=
\dfrac{-21+14i+12i-8i^2}{9-4i^2}
\\\\=
\dfrac{-21+(14+12)i-8i^2}{9-4i^2}
\\\\=
\dfrac{-21+26i-8i^2}{9-4i^2}
.\end{array}
Since $i^2=-1,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{-21+26i-8(-1)}{9-4(-1)}
\\\\=
\dfrac{-21+26i+8}{9+4}
\\\\=
\dfrac{-13+26i}{13}
\\\\=
\dfrac{13(-1+2i)}{13}
\\\\=
\dfrac{\cancel{13}(-1+2i)}{\cancel{13}}
\\\\=
-1+2i
.\end{array}