(2-2i)(3-5i)=0

Solution for (2-2i)(3-5i)=0 equation:

(2-2i)(3-5i)=0

We add all the numbers together, and all the variables

(-2i+2)(-5i+3)=0

We multiply parentheses ..

(+10i^2-6i-10i+6)=0

We get rid of parentheses

10i^2-6i-10i+6=0

We add all the numbers together, and all the variables

10i^2-16i+6=0

a = 10; b = -16; c = +6;
Δ = b2-4ac
Δ = -162-4·10·6
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4}{2*10}=\frac{12}{20}
=3/5
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4}{2*10}=\frac{20}{20}
=1
$