P(x)=-10x^2+500x

Solution for P(x)=-10x^2+500x equation:

(P)=-10P^2+500P

We move all terms to the left:

(P)-(-10P^2+500P)=0

We get rid of parentheses

10P^2-500P+P=0

We add all the numbers together, and all the variables

10P^2-499P=0

a = 10; b = -499; c = 0;
Δ = b2-4ac
Δ = -4992-4·10·0
Δ = 249001
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{249001}=499$
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-499)-499}{2*10}=\frac{0}{20}
=0
$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-499)+499}{2*10}=\frac{998}{20}
=49+9/10
$