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)
 A: Keep it simple:
$$\underbrace{3z^2+2z-5}+(3z+5)i = \underbrace{(3z+5)(z-1)}+(3z+5)i$$
$$=(3z+5)(z-1+i)$$
so $z_1=-5/3$ and $z_2 =1-i$
A: Suppose $a+bi\ne 0$.  Suppose we want to solve $(x+yi)^2 = a+bi$.
Then $(x+yi)^2 =x^2 + 2xyi -y^2 = (x^2 -y^2) + 2xyi = a+bi$.  As $a,b,x^2-y^2, 2xy\in \mathbb R$ we must have
$x^2 - y^2 =a$ and $2xy =b$.
So solve that.
$x = \frac b{2y}$ so $\frac{b^2}{4y^2} - y^2 = a$ so $y^4 +y^2a -\frac 14b^2=0$ so $y^2 = \frac {-1\pm\sqrt {1+ab}}{2}$ ad so on.
So lets solve $(x+yi)^2 =(x^2-y^2) + 2xy = 55 -48i$ so $x^2 - y^2 = 55$ and $2xy =-48$.
$y =-\frac {24}x$ so $x^2 - 24^2 \frac 1{x^2} = 55$ so $x^4 -55x^2 -24^2 = 0$
$x^2 = \frac {55 \pm \sqrt{5^2\cdot 11^2 + 4\cdot 2^2\cdot 12^2}}2=$
yea gods... calculator time ....  $x^2 = \frac {55 \pm 73}2$.  But as $x^2 > 0$ we have $x^2 = \frac {128}2 = 64$.  So $x =\pm 8$.
And $y = \mp \frac {24}8 = \mp 3$.
So the discriminate is $\pm 8 \mp 3i$.
Just like wolfram said.
A: A square root of $55-48i$ is a complex number $a+bi$ such that $(a+bi)^2=55-48i$. In other words, $a^2+2abi-b^2=55-48i$. This is the same thing as asserting that$$\left\{\begin{array}{l}a^2-b^2=55\\2ab=48(\iff ab=24).\end{array}\right.$$It is not necessarily true that there is a solution with $a,b\in\Bbb Z$, but if there is, you could start by looking for two integers $a$ and $b$ whose product is $24$. There aren't many choices. And, indeed, $8\times3=24$ and $8^2-3^2=55$. So, in fact, $8+3i$ is a square root of $55-48i$.
A: 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 $\sqrt{73} \cdot ( \cos(\theta/2) + i \sin(\theta/2) ) = 8 - 3i$.
