Y=-0,02x^2+x

Solution for Y=-0,02x^2+x equation:

=-0.02Y^2+Y

We move all terms to the left:

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

We get rid of parentheses

0.02Y^2-Y=0

We add all the numbers together, and all the variables

0.02Y^2-1Y=0

a = 0.02; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·0.02·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*0.02}=\frac{0}{0.04}
=0
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*0.02}=\frac{2}{0.04}
=50
$