P(x)=-0.1x^2+40x-500

Solution for P(x)=-0.1x^2+40x-500 equation:

(P)=-0.1P^2+40P-500

We move all terms to the left:

(P)-(-0.1P^2+40P-500)=0

We get rid of parentheses

0.1P^2-40P+P+500=0

We add all the numbers together, and all the variables

0.1P^2-39P+500=0

a = 0.1; b = -39; c = +500;
Δ = b2-4ac
Δ = -392-4·0.1·500
Δ = 1321
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-39)-\sqrt{1321}}{2*0.1}=\frac{39-\sqrt{1321}}{0.2}
$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-39)+\sqrt{1321}}{2*0.1}=\frac{39+\sqrt{1321}}{0.2}
$