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

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

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

We add all the numbers together, and all the variables

(5i-4)(-1i+5)=0

We multiply parentheses ..

(-5i^2+25i+4i-20)=0

We get rid of parentheses

-5i^2+25i+4i-20=0

We add all the numbers together, and all the variables

-5i^2+29i-20=0

a = -5; b = 29; c = -20;
Δ = b2-4ac
Δ = 292-4·(-5)·(-20)
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-21}{2*-5}=\frac{-50}{-10}
=+5
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+21}{2*-5}=\frac{-8}{-10}
=4/5
$