Y=-0.002x^2+0.5x

Solution for Y=-0.002x^2+0.5x equation:

=-0.002Y^2+0.5Y

We move all terms to the left:

-(-0.002Y^2+0.5Y)=0

We get rid of parentheses

0.002Y^2-0.5Y=0

a = 0.002; b = -0.5; c = 0;
Δ = b2-4ac
Δ = -0.52-4·0.002·0
Δ = 0.25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-0.5)-\sqrt{0.25}}{2*0.002}=\frac{0.5-\sqrt{0.25}}{0.004}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.5)+\sqrt{0.25}}{2*0.002}=\frac{0.5+\sqrt{0.25}}{0.004}
$