Answer
$\dfrac{6}{5}+\dfrac{8}{5}i$
Work Step by Step
Multiply both the denominator and the numerator by $3+4i$, which is the conjugate of $3-4i$, to obtain:
\begin{align*}
\frac{10}{3-4i}\cdot \frac{3+4i}{3+4i}&=\frac{10(3+4i)}{(3-4i)(3+4i)}\\
\\&=\frac{30+40i}{3^2+4^2} &(\text{note that:}(a-bi)(a+bi)=a^2+b^2)\\
\\&=\frac{30+40i}{9+16} \\
\\&=\frac{30}{25}+\frac{40}{25}i\\
\\&=\frac{6}{5}+\frac{8}{5}i
\end{align*}
