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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

-3i^2+6i-2i+4=0

We add all the numbers together, and all the variables

-3i^2+4i+4=0

a = -3; b = 4; c = +4;
Δ = b2-4ac
Δ = 42-4·(-3)·4
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-8}{2*-3}=\frac{-12}{-6}
=+2
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+8}{2*-3}=\frac{4}{-6}
=-2/3
$