Answer
$\dfrac{-11-7i}{34}$
Work Step by Step
Multiplying by the conjugate of the denominator, the given expression is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{-2+i}{3-5i}
\\\\=
\dfrac{-2+i}{3-5i}\cdot\dfrac{3+5i}{3+5i}
\\\\=
\dfrac{(-2+i)(3+5i)}{(3-5i)(3+5i)}
.\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{(-2+i)(3+5i)}{(3-5i)(3+5i)}
\\\\=
\dfrac{(-2+i)(3+5i)}{(3)^2-(5i)^2}
\\\\=
\dfrac{(-2+i)(3+5i)}{9-25i^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{-2(3)-2(5i)+i(3)+i(5i)}{9-25i^2}
\\\\=
\dfrac{-6-10i+3i+5i^2}{9-25i^2}
\\\\=
\dfrac{-6-7i+5i^2}{9-25i^2}
.\end{array}
Since $i^2=-1,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{-6-7i+5i^2}{9-25i^2}
\\\\=
\dfrac{-6-7i+5(-1)}{9-25(-1)}
\\\\=
\dfrac{-6-7i-5}{9+25}
\\\\=
\dfrac{-11-7i}{34}
.\end{array}
