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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-6i^2+15i-8i+20)=0

We get rid of parentheses

-6i^2+15i-8i+20=0

We add all the numbers together, and all the variables

-6i^2+7i+20=0

a = -6; b = 7; c = +20;
Δ = b2-4ac
Δ = 72-4·(-6)·20
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-23}{2*-6}=\frac{-30}{-12}
=2+1/2
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+23}{2*-6}=\frac{16}{-12}
=-1+1/3
$