Solve $x^2+(-7-4i)x+9+15i=0.$ Solve $$x^2+(-7-4i)x+9+15i=0.$$
Using the quadratic formula, I get $$\frac12 (7+4i \pm \sqrt{-4i})$$ but that's not correct. How do you solve this? I get no help from looking at wolfram alpha.
 A: Given:
$$x^2+(-7-4i)x+(9+15i) = 0$$
We use the well known formula to find
$$\begin{align}
x & = \frac{-b \pm \sqrt{b^2 - 4 \cdot a \cdot c}}{2  \cdot a} \\
  & = \frac{-(-7-4i)\pm \sqrt{(-7-4i)^2 - 4 \cdot 1 \cdot (9+15i)}}{2  \cdot 1} \\
  & = \frac{7+4i \pm\sqrt{(49 + 56i -16) - (36+60i)}}{2} \\
  & = \frac{7+4i \pm\sqrt{-3 - 4i}}{2} \\
\end{align}$$
Now to find $\sqrt{-3 - 4i}$ we note that this can expressed as $r (\cos(\theta)+i\sin(\theta))$ where $r = \sqrt{3^2 +4^2}$ and $\theta = \tan^{-1}\left(\frac{-3}{-4}\right)$.
$\sqrt{r (\cos(\theta)+i\sin(\theta)} = \pm \sqrt{r} \cdot \left( \cos \left(\frac{\theta}{2}\right)  + i \sin \left(\frac{\theta}{2}\right) \right)$
Alternatively if you are not using a calculator we need to find $a + bi$ such that $(a+bi)^2 = -3 -4i$
Note: $(a + bi)^2 = (a^2 - b^2) + 2 \cdot a \cdot b \cdot i$ 
We therefore have two simultaneous equations to solve:
$a^2 - b^2 = -3$ and $2 \cdot a \cdot b = -4 \Rightarrow a \cdot b = -2$
Ignoring the signs we can see that $|b| \gt |a|$ as the the real part is negative and $a$ and $b$ must have opposite signs as the imaginary part is negative. A little trial and error should show you that $(1 - 2i)$ and $(-1 + 2i)$ are solutions to $\sqrt{-3-4i}$
I'm sure you can take it from there.
A: the discriminant in your question is $\sqrt{-3-4i}$
furhter simplify the term inside the square root in the form
$$r(\cos\theta+i\sin\theta)$$
taking square root changes it to
$$\sqrt{r}(\cos\frac{\theta}{2}+i\sin\frac{\theta}{2})$$
A: Firstly you have used the quadratic formula incorrectly.
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2}$$
gives,
$$x=\frac{7+4i\pm\sqrt{(7+4i)^2-36-60i}}{2}$$
$$x=\frac{7+4i\pm\sqrt{-3-4i}}{2}$$
Now try to find the value of $\sqrt{-3-4i}$. Let, $a+bi=\sqrt{-3-4i}$ and try to solve this to find the values of $a$ and $b$. 
