P(x)=0.5x^2+40x-300

Solution for P(x)=0.5x^2+40x-300 equation:

(P)=0.5P^2+40P-300

We move all terms to the left:

(P)-(0.5P^2+40P-300)=0

We get rid of parentheses

-0.5P^2+P-40P+300=0

We add all the numbers together, and all the variables

-0.5P^2-39P+300=0

a = -0.5; b = -39; c = +300;
Δ = b2-4ac
Δ = -392-4·(-0.5)·300
Δ = 2121
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{2121}}{2*-0.5}=\frac{39-\sqrt{2121}}{-1}
$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-39)+\sqrt{2121}}{2*-0.5}=\frac{39+\sqrt{2121}}{-1}
$