9=0.005x^2+0.002x

Solution for 9=0.005x^2+0.002x equation:

9=0.005x^2+0.002x

We move all terms to the left:

9-(0.005x^2+0.002x)=0

We get rid of parentheses

-0.005x^2-0.002x+9=0

a = -0.005; b = -0.002; c = +9;
Δ = b2-4ac
Δ = -0.0022-4·(-0.005)·9
Δ = 0.180004
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-0.002)-\sqrt{0.180004}}{2*-0.005}=\frac{0.002-\sqrt{0.180004}}{-0.01}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.002)+\sqrt{0.180004}}{2*-0.005}=\frac{0.002+\sqrt{0.180004}}{-0.01}
$