-0,5x^2+x=0

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Solution for -0,5x^2+x=0 equation: 



-0.5x^2+x=0

a = -0.5; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-0.5)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$


$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*-0.5}=\frac{-2}{-1}
=+2
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-0.5}=\frac{0}{-1}
=0
$

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