p^2+40p-3696=0

Solution for p^2+40p-3696=0 equation:

p^2+40p-3696=0

a = 1; b = 40; c = -3696;
Δ = b2-4ac
Δ = 402-4·1·(-3696)
Δ = 16384
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16384}=128$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-128}{2*1}=\frac{-168}{2}
=-84
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+128}{2*1}=\frac{88}{2}
=44
$