P=-8n2+40n-3

Solution for P=-8n2+40n-3 equation:

=-8P^2+40P-3

We move all terms to the left:

-(-8P^2+40P-3)=0

We get rid of parentheses

8P^2-40P+3=0

a = 8; b = -40; c = +3;
Δ = b2-4ac
Δ = -402-4·8·3
Δ = 1504
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:

$\sqrt{\Delta}=\sqrt{1504}=\sqrt{16*94}=\sqrt{16}*\sqrt{94}=4\sqrt{94}$
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-4\sqrt{94}}{2*8}=\frac{40-4\sqrt{94}}{16}
$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+4\sqrt{94}}{2*8}=\frac{40+4\sqrt{94}}{16}
$