#### Answer

$\dfrac{8+12i}{13}$

#### Work Step by Step

Multiplying by the conjugate of the denominator, the given expression, $
\dfrac{4}{2-3i}
,$ is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{4}{2-3i}\cdot\dfrac{2+3i}{2+3i}
\\\\=
\dfrac{4(2)+4(3i)}{(2)^2-(3i)^2}
\\\\=
\dfrac{8+12i}{4-9i^2}
\\\\=
\dfrac{8+12i}{4-9(-1)}
\\\\=
\dfrac{8+12i}{4+9}
\\\\=
\dfrac{8+12i}{13}
.\end{array}