0=(-3-2i)*(-3+2i)

Solution for 0=(-3-2i)*(-3+2i) equation:

0=(-3-2i)(-3+2i)

We move all terms to the left:

0-((-3-2i)(-3+2i))=0

We add all the numbers together, and all the variables

-((-2i-3)(2i-3))+0=0

We add all the numbers together, and all the variables

-((-2i-3)(2i-3))=0

We multiply parentheses ..

-((-4i^2+6i-6i+9))=0


We calculate terms in parentheses: -((-4i^2+6i-6i+9)), so:

(-4i^2+6i-6i+9)

We get rid of parentheses

-4i^2+6i-6i+9

We add all the numbers together, and all the variables

-4i^2+9

Back to the equation:

-(-4i^2+9)


We get rid of parentheses

4i^2-9=0

a = 4; b = 0; c = -9;
Δ = b2-4ac
Δ = 02-4·4·(-9)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12}{2*4}=\frac{-12}{8}
=-1+1/2
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12}{2*4}=\frac{12}{8}
=1+1/2
$