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)

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.
