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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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

a = -15; b = -2; c = +24;
Δ = b2-4ac
Δ = -22-4·(-15)·24
Δ = 1444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1444}=38$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-38}{2*-15}=\frac{-36}{-30}
=1+1/5
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+38}{2*-15}=\frac{40}{-30}
=-1+1/3
$