(6-3i)(4+2i)=0

Solution for (6-3i)(4+2i)=0 equation:

(6-3i)(4+2i)=0

We add all the numbers together, and all the variables

(-3i+6)(2i+4)=0

We multiply parentheses ..

(-6i^2-12i+12i+24)=0

We get rid of parentheses

-6i^2-12i+12i+24=0

We add all the numbers together, and all the variables

-6i^2+24=0

a = -6; b = 0; c = +24;
Δ = b2-4ac
Δ = 02-4·(-6)·24
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24}{2*-6}=\frac{-24}{-12}
=+2
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24}{2*-6}=\frac{24}{-12}
=-2
$