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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

-6i^2-4i-9i-6=0

We add all the numbers together, and all the variables

-6i^2-13i-6=0

a = -6; b = -13; c = -6;
Δ = b2-4ac
Δ = -132-4·(-6)·(-6)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-5}{2*-6}=\frac{8}{-12}
=-2/3
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+5}{2*-6}=\frac{18}{-12}
=-1+1/2
$