-2v^2+2v=0

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Solution for -2v^2+2v=0 equation: 



-2v^2+2v=0

a = -2; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·(-2)·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$


$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*-2}=\frac{-4}{-4}
=1
$


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

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