(4+3i)(5-6i)=0

Solution for (4+3i)(5-6i)=0 equation:

(4+3i)(5-6i)=0

We add all the numbers together, and all the variables

(3i+4)(-6i+5)=0

We multiply parentheses ..

(-18i^2+15i-24i+20)=0

We get rid of parentheses

-18i^2+15i-24i+20=0

We add all the numbers together, and all the variables

-18i^2-9i+20=0

a = -18; b = -9; c = +20;
Δ = b2-4ac
Δ = -92-4·(-18)·20
Δ = 1521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1521}=39$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-39}{2*-18}=\frac{-30}{-36}
=5/6
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+39}{2*-18}=\frac{48}{-36}
=-1+1/3
$