P=-0.5x^2+40x-740

Solution for P=-0.5x^2+40x-740 equation:

=-0.5P^2+40P-740

We move all terms to the left:

-(-0.5P^2+40P-740)=0

We get rid of parentheses

0.5P^2-40P+740=0

a = 0.5; b = -40; c = +740;
Δ = b2-4ac
Δ = -402-4·0.5·740
Δ = 120
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{120}=\sqrt{4*30}=\sqrt{4}*\sqrt{30}=2\sqrt{30}$
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-2\sqrt{30}}{2*0.5}=\frac{40-2\sqrt{30}}{1}
$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+2\sqrt{30}}{2*0.5}=\frac{40+2\sqrt{30}}{1}
$