<s>[INST] Is my result to $z^2 - (3+4i)z - 1 + 7i = 0$ right? I solved it much like I would a second degree equation, with a=1, b=-(3+4i) and c=(-1+7i). Is that the correct approach?
My final result was $z = \frac{3+4i \pm i\sqrt{3+4i}}{2}$
Is that correct? There is no way for me to check that.
 [/INST] Here is the standard way t o compute the square roots of a complex number: $$(x+iy)^2=3+4i\iff \begin{cases}x^2-y^2=3,\\2xy=4.\end{cases}$$ 
Further the square of the module of $x+iy$ is equal to the module of $3+4i$:
$$x^2+y^2=5.$$
So we first solve the linear system in $x^2$ and $y^2$:
$$\begin{cases}x^2-y^2=3,\\x^2+y^2=5\end{cases}\iff\begin{cases}x^2=4,\\y^2=1\end{cases}\iff\begin{cases}x=\pm2,\\y^2=\pm1.\end{cases}$$
Further,$xy=2>0$ implies $x$ and $y$ have the same sign, and finally we obtain
$$x+iy=\pm(2+i).$$
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