n^2+32n=0

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Solution for n^2+32n=0 equation: 



n^2+32n=0

a = 1; b = 32; c = 0;
Δ = b2-4ac
Δ = 322-4·1·0
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$


$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-32}{2*1}=\frac{-64}{2}
=-32
$


$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+32}{2*1}=\frac{0}{2}
=0
$

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