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

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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 $

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