p^2+8p-128=0

Solution for p^2+8p-128=0 equation:

p^2+8p-128=0

a = 1; b = 8; c = -128;
Δ = b2-4ac
Δ = 82-4·1·(-128)
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-24}{2*1}=\frac{-32}{2}
=-16
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+24}{2*1}=\frac{16}{2}
=8
$