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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-20i^2-12i+10i+6)=0

We get rid of parentheses

-20i^2-12i+10i+6=0

We add all the numbers together, and all the variables

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

a = -20; b = -2; c = +6;
Δ = b2-4ac
Δ = -22-4·(-20)·6
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*-20}=\frac{-20}{-40}
=1/2
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+22}{2*-20}=\frac{24}{-40}
=-3/5
$