p^2+8p=16=0

Solution for p^2+8p=16=0 equation:

p^2+8p=16=0

We move all terms to the left:

p^2+8p-(16)=0

a = 1; b = 8; c = -16;
Δ = b2-4ac
Δ = 82-4·1·(-16)
Δ = 128
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:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{2}}{2*1}=\frac{-8-8\sqrt{2}}{2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{2}}{2*1}=\frac{-8+8\sqrt{2}}{2}
$