P=2x^2+40x-72

Solution for P=2x^2+40x-72 equation:

=2P^2+40P-72

We move all terms to the left:

-(2P^2+40P-72)=0

We get rid of parentheses

-2P^2-40P+72=0

a = -2; b = -40; c = +72;
Δ = b2-4ac
Δ = -402-4·(-2)·72
Δ = 2176
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:

$\sqrt{\Delta}=\sqrt{2176}=\sqrt{64*34}=\sqrt{64}*\sqrt{34}=8\sqrt{34}$
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-8\sqrt{34}}{2*-2}=\frac{40-8\sqrt{34}}{-4}
$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+8\sqrt{34}}{2*-2}=\frac{40+8\sqrt{34}}{-4}
$