P(x)=-0.5x^2+10x+1200

Solution for P(x)=-0.5x^2+10x+1200 equation:

(P)=-0.5P^2+10P+1200

We move all terms to the left:

(P)-(-0.5P^2+10P+1200)=0

We get rid of parentheses

0.5P^2-10P+P-1200=0

We add all the numbers together, and all the variables

0.5P^2-9P-1200=0

a = 0.5; b = -9; c = -1200;
Δ = b2-4ac
Δ = -92-4·0.5·(-1200)
Δ = 2481
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{-(-9)-\sqrt{2481}}{2*0.5}=\frac{9-\sqrt{2481}}{1}
$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{2481}}{2*0.5}=\frac{9+\sqrt{2481}}{1}
$