Y=0.02x^2+9x

Solution for Y=0.02x^2+9x equation:

=0.02Y^2+9Y

We move all terms to the left:

-(0.02Y^2+9Y)=0

We get rid of parentheses

-0.02Y^2-9Y=0

a = -0.02; b = -9; c = 0;
Δ = b2-4ac
Δ = -92-4·(-0.02)·0
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-9}{2*-0.02}=\frac{0}{-0.04}
=0
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+9}{2*-0.02}=\frac{18}{-0.04}
=-450
$