<s>[INST] Find all the solutions of $z^2-(1+3i)z-8-i=0$ I am stuck on a problem and I was hoping someone could tell me what I am doing wrong.

I want to find all the roots of:
$$z^2-(1+3i)z-8-i=0$$

There are two ways I tried to approach this.

*

*Quadratic formula:
$$\begin{align} z_1,z_2 &=-\frac{p}{2}\pm\sqrt{\left( \frac{p}{2}\right)^2-q} \\ & \implies z_1,z_2=\frac{1+3i}{2}\pm\sqrt{\left( \frac{(-1-3i)}{2}\right)^2-(-8-i)}\\ &=  \frac{1+3i}{2} \pm\sqrt{\frac{24+10i}{4} }=\frac{1+3i \pm\sqrt{24+10i}}{2}\end{align}$$


*Completing the square:
$$\begin{aligned} z^2-(1+3i)z-8-i &=0 \\ & \iff \left(z-\left( \frac{1+3i}{2}\right) \right)^2-8-i=\left( \frac{1+3i}{2}\right)^2 \\ & \iff u^2=\frac{24+10i}{4} \\ & \iff u=\pm\frac{\sqrt{24+10i}}{2} \\ & \iff z_{1,2}=\frac{1+3i \pm\sqrt{24+10i}}{2}\end{aligned}  $$
Using both methods I arrive at the same result. However, wolframalpha tells me the roots are:
$$z_1=-2+i \\ z_2=3+2i$$
I have tried everything (writing it in in exponential form, graphing, etc.) but I have no idea how I can arrive at those two roots. What am I doing wrong?
 [/INST] Hint:  Expanding $$(5+i)^2=25+10i-1$$
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