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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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

a = -24; b = 2; c = +15;
Δ = b2-4ac
Δ = 22-4·(-24)·15
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*-24}=\frac{-40}{-48}
=5/6
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+38}{2*-24}=\frac{36}{-48}
=-3/4
$