4n^2+36n=0

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Solution for 4n^2+36n=0 equation: 



4n^2+36n=0

a = 4; b = 36; c = 0;
Δ = b2-4ac
Δ = 362-4·4·0
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$


$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-36}{2*4}=\frac{-72}{8}
=-9
$


$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+36}{2*4}=\frac{0}{8}
=0
$

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