P(x)=-25x^2+500

Solution for P(x)=-25x^2+500 equation:

(P)=-25P^2+500

We move all terms to the left:

(P)-(-25P^2+500)=0

We get rid of parentheses

25P^2+P-500=0

a = 25; b = 1; c = -500;
Δ = b2-4ac
Δ = 12-4·25·(-500)
Δ = 50001
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{-(1)-\sqrt{50001}}{2*25}=\frac{-1-\sqrt{50001}}{50}
$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{50001}}{2*25}=\frac{-1+\sqrt{50001}}{50}
$