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

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

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

We add all the numbers together, and all the variables

(3i+6)(-1i+2)=0

We multiply parentheses ..

(-3i^2+6i-6i+12)=0

We get rid of parentheses

-3i^2+6i-6i+12=0

We add all the numbers together, and all the variables

-3i^2+12=0

a = -3; b = 0; c = +12;
Δ = b2-4ac
Δ = 02-4·(-3)·12
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*-3}=\frac{-12}{-6}
=+2
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12}{2*-3}=\frac{12}{-6}
=-2
$