Y=-0.01x^2+0.3x+2

Solution for Y=-0.01x^2+0.3x+2 equation:

=-0.01Y^2+0.3Y+2

We move all terms to the left:

-(-0.01Y^2+0.3Y+2)=0

We get rid of parentheses

0.01Y^2-0.3Y-2=0

a = 0.01; b = -0.3; c = -2;
Δ = b2-4ac
Δ = -0.32-4·0.01·(-2)
Δ = 0.17
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{0.17}}{2*0.01}=\frac{0.3-\sqrt{0.17}}{0.02}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.3)+\sqrt{0.17}}{2*0.01}=\frac{0.3+\sqrt{0.17}}{0.02}
$