(2-2i)(1+3i)=0

Solution for (2-2i)(1+3i)=0 equation:

(2-2i)(1+3i)=0

We add all the numbers together, and all the variables

(-2i+2)(3i+1)=0

We multiply parentheses ..

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

We get rid of parentheses

-6i^2-2i+6i+2=0

We add all the numbers together, and all the variables

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

a = -6; b = 4; c = +2;
Δ = b2-4ac
Δ = 42-4·(-6)·2
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-8}{2*-6}=\frac{-12}{-12}
=1
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+8}{2*-6}=\frac{4}{-12}
=-1/3
$