<s>[INST] Solving $3z^2 + (2+3i)z + (5i-5) = 0$ In a Precalculus book, Complex numbers chapter, I am given the following exercise:

Solve the equation $3z^2 + (2+3i)z + (5i-5) = 0$, giving your answers in the form $z = a + bi$

From the way that the exercise is shown, I thought about using the quadratic formula to solve it. Starting with the discriminant, I found
$$\Delta = (2+3i)^2 - 4 \cdot 3 \cdot (5i-5) = 55-48i$$
If I could find the root of that complex number, I could finish this exercise. But how could I do that? (Wolfram gives be $8-3i$ as an answer for that root)
(by the way, this exercise comes before the root finding method for complex numbers)
 [/INST] Use de Moivre's formula: $\Delta = 55 - 48i$ is the complex number with magnitude $\sqrt{55^2 + (-48)^2} = 73$ and whose argument $\theta$ satisfies $\cos \theta = \frac{55}{73}, \sin\theta = -\frac{48}{73}$.
Therefore, $\sqrt{\Delta}$ is the complex number with magnitude $\sqrt{73}$, and whose argument is either $\frac{1}{2} \theta$ or $\frac{1}{2} \theta + \pi$.  Now, since $\theta$ is in the fourth quadrant, if we take the choice of $\theta$ where $-\frac{\pi}{2} < \theta < 0$, then $\frac{1}{2} \theta$ is also in the fourth quadrant.  Now,
$$\cos\left(\frac{1}{2} \theta\right) = \sqrt{\frac{1 + \cos{\theta}}{2}} = \sqrt{\frac{64}{73}} = \frac{8}{\sqrt{73}}$$
and
$$\sin\left(\frac{1}{2} \theta\right) = -\sqrt{\frac{1 - \cos{\theta}}{2}} = -\sqrt{\frac{9}{73}} = -\frac{3}{\sqrt{73}}.$$
Therefore, one of the square roots is $\sqrt{73} \cdot ( \cos(\theta/2) + i \sin(\theta/2) ) = 8 - 3i$.
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