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.
 A: 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\\
$
$\cos\theta = \frac 35\\
\cos\frac{\theta}2 = \sqrt {\frac {1+\frac 35}{2}} = \frac {2}{\sqrt 5}\\
\sin\frac{\theta}2 =-\sqrt {\frac {1-\frac 35}{2}} = -\frac 1{\sqrt 5}$
$\sqrt {3 +4i} = 2 +i$
$z = \frac {(3+4i) \pm i(2+i)}2$
$z = 2+i,1+3i$
A: Checking the roots is in principle just a matter of arithmetic. But it would be very easy to make a mistake with all those $i$'s and square roots.
An easier computation is to check that the sum and product of the two roots is correct. Given the equation $z^2+bx+c=0$, the sum of the roots is $-b$ and the product of the roots  is $c$.
In this case, the sum of your two proposed roots is clearly $3+4i$, so that checks out.
So now you have to check the product: i.e. you have to check that
$$\frac14\{(3+4i)^2-(i\sqrt{3+4i})^2\} = -1+7i$$
Evaluating the two squared terms:
$$\frac14\{(-7+24i)+(3+4i)\} = -1+7i$$
By my reckoning, this equation is correct. So those are indeed the correct roots.
A: Solving it using the quadratic formula is the way to go:
$$z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$$
where your equation is $az^2 + bz + c = 0.$
You can always check your answer by substituting the solution back into the equation.  But your value of $b$ is actually $-(3+4i)$ so your answers aren't correct if you implemented the quadratic formula correctly.  (It may just as easily be a typo in your question.)
A: 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
$$x+iy=\pm(2+i).$$
