-0,5x^2+0,5x=0

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

-0.5x^2+0.5x=0

a = -0.5; b = 0.5; c = 0;
Δ = b2-4ac
Δ = 0.52-4·(-0.5)·0
Δ = 0.25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0.5)-\sqrt{0.25}}{2*-0.5}=\frac{-0.5-\sqrt{0.25}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.5)+\sqrt{0.25}}{2*-0.5}=\frac{-0.5+\sqrt{0.25}}{-1}
$