P(x)=230+40x-0.5x^2

Solution for P(x)=230+40x-0.5x^2 equation:

(P)=230+40P-0.5P^2

We move all terms to the left:

(P)-(230+40P-0.5P^2)=0

We get rid of parentheses

0.5P^2-40P+P-230=0

We add all the numbers together, and all the variables

0.5P^2-39P-230=0

a = 0.5; b = -39; c = -230;
Δ = b2-4ac
Δ = -392-4·0.5·(-230)
Δ = 1981
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{1981}}{2*0.5}=\frac{39-\sqrt{1981}}{1}
$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-39)+\sqrt{1981}}{2*0.5}=\frac{39+\sqrt{1981}}{1}
$