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.

You are not done yet.  There is no reason to leave complex numbers under the radical.
$z = \frac {(3+4i) \pm i\sqrt {3 + 4i}}2$
$3 + 4i = 5 e^{i\theta}\\
\sqrt {+3 +4i} = \sqrt 5 e^{i\frac {\theta}{2}}$
$e^{i\theta} = \cos\theta + i \sin\theta\\
e^{i\frac{\theta}2} = \cos\frac{\theta}2 + i \sin\frac{\theta}2\\
$
$\cos\theta = \frac 35\\
\cos\frac{\theta}2 = \sqrt {\frac {1+\frac 35}{2}} = \frac {2}{\sqrt 5}\\
\sin\frac{\theta}2 =-\sqrt {\frac {1-\frac 35}{2}} = -\frac 1{\sqrt 5}$
$\sqrt {3 +4i} = 2 +i$
$z = \frac {(3+4i) \pm i(2+i)}2$
$z = 2+i,1+3i$
