(3-2i)(5+4i)=0

Solution for (3-2i)(5+4i)=0 equation:

(3-2i)(5+4i)=0

We add all the numbers together, and all the variables

(-2i+3)(4i+5)=0

We multiply parentheses ..

(-8i^2-10i+12i+15)=0

We get rid of parentheses

-8i^2-10i+12i+15=0

We add all the numbers together, and all the variables

-8i^2+2i+15=0

a = -8; b = 2; c = +15;
Δ = b2-4ac
Δ = 22-4·(-8)·15
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-22}{2*-8}=\frac{-24}{-16}
=1+1/2
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+22}{2*-8}=\frac{20}{-16}
=-1+1/4
$