-2y^2+10y=0

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Solution for -2y^2+10y=0 equation: 



-2y^2+10y=0

a = -2; b = 10; c = 0;
Δ = b2-4ac
Δ = 102-4·(-2)·0
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$


$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10}{2*-2}=\frac{-20}{-4}
=+5
$


$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10}{2*-2}=\frac{0}{-4}
=0
$

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