Solve a complex equation Solve the following equation $$(4-3i)z^2-25z+31-17i= 0 $$
Dividing by 4-3i gives me $$z^2  \frac{-100z-75zi + 124 + 93i -68i -51i^2}{25}$$ 
which goes to $$z^2 -4z-3zi + 7+i$$
then i collect the terms so $$z - \left(\frac{(4-3i)}{2}\right)^2 = -7 -i + \left(\frac{4-3i}{2}\right)^2$$
and after that i can't get the expansion to work.
Can you help me out? 

Notice, $$(4-3i)z^2-25z+31-17i=0$$ Solving the above quadratic equation for $z$ as follows $$z=\frac{-(-25)\pm\sqrt{(-25)^2-4(4-3i)(31-17i)}}{2(4-3i)}$$
$$z=\frac{-(-25)\pm\sqrt{(-25)^2-4(4-3i)(31-17i)}}{2(4-3i)}$$ $$z=\frac{(4+3i)(25\pm\sqrt{333+644i})}{2(16+9)}$$
$$z=\frac{(4+3i)(25\pm\sqrt{333+644i})}{50}$$
$$z=\frac{(4+3i)\left(25\pm\sqrt{725}\left(\cos \frac{\alpha}{2}+i\sin\frac{\alpha}{2}\right)\right)}{50}$$ Where, $\alpha=\cos^{-1}\left(\frac{333}{725}\right)=\sin^{-1}\left(\frac{644}{725}\right)$
I hope you can take it from here.
