Y=-0.02x^2+2x

Solution for Y=-0.02x^2+2x equation:

=-0.02Y^2+2Y

We move all terms to the left:

-(-0.02Y^2+2Y)=0

We get rid of parentheses

0.02Y^2-2Y=0

a = 0.02; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·0.02·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*0.02}=\frac{0}{0.04}
=0
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*0.02}=\frac{4}{0.04}
=100
$