(g^2)+40g=0

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Solution for (g^2)+40g=0 equation: 



(g^2)+40g=0

a = 1; b = 40; c = 0;
Δ = b2-4ac
Δ = 402-4·1·0
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$


$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-40}{2*1}=\frac{-80}{2}
=-40
$


$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+40}{2*1}=\frac{0}{2}
=0
$

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