Answer
$-2+i$
Work Step by Step
Step 1: Multiplying both the numerator and the denominator of the expression by the complex conjugate of the denominator:
$\frac{-3+4i}{2-i}\times\frac{2+i}{2+i}$
Step 2: $\frac{-3+4i}{2-i}\times\frac{2+i}{2+i}=\frac{(-3+4i)(2+i)}{(2)^{2}-(i)^{2}}$
Step 3: $\frac{(-3+4i)(2+i)}{(2)^{2}-(i)^{2}}=\frac{-6-3i+8i+4i^{2}}{4-(-1)}=\frac{-6+5i+4(-1)}{5}$
Step 4: $\frac{-6+5i+4(-1)}{5}=\frac{-10+5i}{5}=\frac{5(-2+i)}{2}=-2+i$