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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-4i^2+16i-5i+20)=0

We get rid of parentheses

-4i^2+16i-5i+20=0

We add all the numbers together, and all the variables

-4i^2+11i+20=0

a = -4; b = 11; c = +20;
Δ = b2-4ac
Δ = 112-4·(-4)·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{-(11)-21}{2*-4}=\frac{-32}{-8}
=+4
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+21}{2*-4}=\frac{10}{-8}
=-1+1/4
$