(6-2i)(3+i)=0

Solution for (6-2i)(3+i)=0 equation:

(6-2i)(3+i)=0

We add all the numbers together, and all the variables

(-2i+6)(i+3)=0

We multiply parentheses ..

(-2i^2-6i+6i+18)=0

We get rid of parentheses

-2i^2-6i+6i+18=0

We add all the numbers together, and all the variables

-2i^2+18=0

a = -2; b = 0; c = +18;
Δ = b2-4ac
Δ = 02-4·(-2)·18
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{-(0)-12}{2*-2}=\frac{-12}{-4}
=+3
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12}{2*-2}=\frac{12}{-4}
=-3
$