-10x^2+40x=0

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Solution for -10x^2+40x=0 equation:



-10x^2+40x=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:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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

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