Y=-0.005x^2+0.3x

Solution for Y=-0.005x^2+0.3x equation:

=-0.005Y^2+0.3Y

We move all terms to the left:

-(-0.005Y^2+0.3Y)=0

We get rid of parentheses

0.005Y^2-0.3Y=0

a = 0.005; b = -0.3; c = 0;
Δ = b2-4ac
Δ = -0.32-4·0.005·0
Δ = 0.09
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{-(-0.3)-\sqrt{0.09}}{2*0.005}=\frac{0.3-\sqrt{0.09}}{0.01}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.3)+\sqrt{0.09}}{2*0.005}=\frac{0.3+\sqrt{0.09}}{0.01}
$