10w2+40w=0

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Solution for 10w2+40w=0 equation:



10w^2+40w=0
a = 10; b = 40; c = 0;
Δ = b2-4ac
Δ = 402-4·10·0
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-40}{2*10}=\frac{-80}{20} =-4 $
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+40}{2*10}=\frac{0}{20} =0 $

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