To divide two complex numbers, I need to multiply both the numerator and the denominator by the conjugate of the denominator, which is the same number but with the opposite sign of the imaginary part. This will make the denominator a real number and simplify the fraction. The conjugate of $1+2i$ is $1-2i$. So, I need to multiply $\frac{-3+4i}{1+2i}$ by $\frac{1-2i}{1-2i}$. Using the distributive property, I get $\frac{-3+4i}{1+2i}\cdot\frac{1-2i}{1-2i}=\frac{(-3+4i)(1-2i)}{(1+2i)(1-2i)}$. Expanding the brackets, I get $\frac{-3+6i+4i-8i^2}{1-2i+2i-4i^2}$. Simplifying, I get $\frac{-3+10i+8}{1+4}=\frac{5+10i}{5}$. Dividing both the real and the imaginary part by 5, I get $\frac{5+10i}{5}=\frac{5}{5}+\frac{10}{5}i=1+2i$. 
William Sydney Porter tried to perform the calculation $\frac{-3+4i}{1+2i}$. However, he accidentally missed the minus sign, finding $\frac{3+4i}{1+2i}=\frac{11}{5}-\frac{2}{5}i$. What answer should he have obtained?