#### Answer

$\dfrac{1-12i}{15}$

#### Work Step by Step

Multiplying by the conjugate of the denominator, the given expression, $
\dfrac{5-2i}{3+6i}
,$ is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{5-2i}{3+6i}\cdot\dfrac{3-6i}{3-6i}
\\\\=
\dfrac{5(3)+5(-6i)-2i(3)-2i(-6i)}{(3)^2-(6i)^2}
\\\\=
\dfrac{15-30i-6i+12i^2}{9-36i^2}
\\\\=
\dfrac{15-30i-6i+12(-1)}{9-36(-1)}
\\\\=
\dfrac{15-30i-6i-12}{9+36}
\\\\=
\dfrac{(15-12)+(-30i-6i)}{9+36}
\\\\=
\dfrac{3-36i}{45}
\\\\=
\dfrac{3(1-12i)}{45}
\\\\=
\dfrac{\cancel{3}(1-12i)}{\cancel{3}\cdot15}
\\\\=
\dfrac{1-12i}{15}
\end{array}