(2-2i)(3+4i)=0

Solution for (2-2i)(3+4i)=0 equation:

(2-2i)(3+4i)=0

We add all the numbers together, and all the variables

(-2i+2)(4i+3)=0

We multiply parentheses ..

(-8i^2-6i+8i+6)=0

We get rid of parentheses

-8i^2-6i+8i+6=0

We add all the numbers together, and all the variables

-8i^2+2i+6=0

a = -8; b = 2; c = +6;
Δ = b2-4ac
Δ = 22-4·(-8)·6
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-14}{2*-8}=\frac{-16}{-16}
=1
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+14}{2*-8}=\frac{12}{-16}
=-3/4
$