4n^2+20n=0

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Solution for 4n^2+20n=0 equation: 



4n^2+20n=0

a = 4; b = 20; c = 0;
Δ = b2-4ac
Δ = 202-4·4·0
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$


$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20}{2*4}=\frac{-40}{8}
=-5
$


$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20}{2*4}=\frac{0}{8}
=0
$

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