#### Answer

$\dfrac{-3+4i}{2-i}=-2+i$

#### Work Step by Step

$\dfrac{-3+4i}{2-i}$
Begin the evaluation of the quotient by multiplying the numerator and the denominator by the complex conjugate of the denominator:
$\dfrac{-3+4i}{2-i}\cdot\dfrac{2+i}{2+i}=\dfrac{(-3+4i)(2+i)}{2^{2}-i^{2}}=...$
Evaluate the operations indicated in the numerator and in the denominator:
$...=\dfrac{-6-3i+8i+4i^{2}}{4-i^{2}}=...$
Substitute $i^{2}$ by $−1$ and simplify:
$...=\dfrac{-6-3i+8i+4(-1)}{4-(-1)}=\dfrac{-6+5i-4}{4+1}=\dfrac{-10+5i}{5}=...$
$...=-\dfrac{10}{5}+\dfrac{5}{5}i=-2+i$