(2+i)(2+4i)=0

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



(2+i)(2+4i)=0
We add all the numbers together, and all the variables
(i+2)(4i+2)=0
We multiply parentheses ..
(+4i^2+2i+8i+4)=0
We get rid of parentheses
4i^2+2i+8i+4=0
We add all the numbers together, and all the variables
4i^2+10i+4=0
a = 4; b = 10; c = +4;
Δ = b2-4ac
Δ = 102-4·4·4
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-6}{2*4}=\frac{-16}{8} =-2 $
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+6}{2*4}=\frac{-4}{8} =-1/2 $

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