(-3-3i)(-4+3i)=0

Solution for (-3-3i)(-4+3i)=0 equation:

(-3-3i)(-4+3i)=0

We add all the numbers together, and all the variables

(-3i-3)(3i-4)=0

We multiply parentheses ..

(-9i^2+12i-9i+12)=0

We get rid of parentheses

-9i^2+12i-9i+12=0

We add all the numbers together, and all the variables

-9i^2+3i+12=0

a = -9; b = 3; c = +12;
Δ = b2-4ac
Δ = 32-4·(-9)·12
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-21}{2*-9}=\frac{-24}{-18}
=1+1/3
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+21}{2*-9}=\frac{18}{-18}
=-1
$