0=-0.0002x^2+40x

Solution for 0=-0.0002x^2+40x equation:

0=-0.0002x^2+40x

We move all terms to the left:

0-(-0.0002x^2+40x)=0

We add all the numbers together, and all the variables

-(-0.0002x^2+40x)=0

We get rid of parentheses

0.0002x^2-40x=0

a = 0.0002; b = -40; c = 0;
Δ = b2-4ac
Δ = -402-4·0.0002·0
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-40}{2*0.0002}=\frac{0}{0.0004}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+40}{2*0.0002}=\frac{80}{0.0004}
=200000
$