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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-6i^2+12i-4i+8)=0

We get rid of parentheses

-6i^2+12i-4i+8=0

We add all the numbers together, and all the variables

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

a = -6; b = 8; c = +8;
Δ = b2-4ac
Δ = 82-4·(-6)·8
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{-(8)-16}{2*-6}=\frac{-24}{-12}
=+2
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+16}{2*-6}=\frac{8}{-12}
=-2/3
$