Solution to some confusing complex equation

Question

I am asked to solve the equation:

$(4-3i)z^2-25z+31-17i = 0$

I approached it this way:

$(4-3i)$[$z^2$-$(25z/((4-3i))$+$(31/(4-3i))$-$(17i/(4-3i))$]=0 which led to

$z^2$-$(25z/((4-3i))$+$(31/(4-3i))$-$(17i/(4-3i))$=0
which I simplified by realising all the denominators to eventually get

$z^2-z(4+3i)+7+i=0$

and this is the point where I seem to get lost. I tried to complete the square on the above equation but that gave some weird solutions, then I tried using [$Z^2$-(sum of roots)z + (product of roots)], and still this did not lead anywhere. I tried working in polar form, but this also gave me some unsightly solutions.

The solutions I seek are $z=3+4i$ and $z=1-i$. I wonder if there is anywhere I am making a mistake, but any contribution will be highly appreciated!

Answer 1

$$z^2-z(4+3i)+7+i=0\implies z=\frac{(4+3i)\pm\sqrt{(4+3i)^2-4(7+i)}}2$$

Now, $$(4+3i)^2-4(7+i)=-21+20i=2\cdot5i\cdot2+(5i)^2+2^2=(5i+2)^2$$

Answer 2

Hint: Use the quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2}$.