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

Solution for (2-i)(8+2i)=0 equation:

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

We add all the numbers together, and all the variables

(-1i+2)(2i+8)=0

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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

a = -2; b = -4; c = +16;
Δ = b2-4ac
Δ = -42-4·(-2)·16
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-12}{2*-2}=\frac{-8}{-4}
=+2
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+12}{2*-2}=\frac{16}{-4}
=-4
$