complex equation with non-zero $z^2$

Question

$$ (2+i)z^2+(1-7i)z-5 = 0 $$ I started with dividing with $(2+i)$ and got $z^2-z(1-3i)-(2-i)$. Now I'm trying to completing the square but fails. I get $$ \left(z-\frac{1+3i}{2}\right)^2-\left(\frac{-1+3i}{2}\right)^2-2+i = 0 $$

Answer 1

Use the quadratic formula, and compute the square root of the resulting discriminant either by inspection or letting $\sqrt{\Delta}=a+bi$ and then solving for $a$ and $b$ by squaring both sides and equating real/imaginary components.