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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-4i^2-8i+8i+16)=0

We get rid of parentheses

-4i^2-8i+8i+16=0

We add all the numbers together, and all the variables

-4i^2+16=0

a = -4; b = 0; c = +16;
Δ = b2-4ac
Δ = 02-4·(-4)·16
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{-(0)-16}{2*-4}=\frac{-16}{-8}
=+2
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16}{2*-4}=\frac{16}{-8}
=-2
$