(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^2-4i+6=0

a = -10; b = -4; c = +6;
Δ = b2-4ac
Δ = -42-4·(-10)·6
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-16}{2*-10}=\frac{-12}{-20}
=3/5
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+16}{2*-10}=\frac{20}{-20}
=-1
$