p^2+8=104

Solution for p^2+8=104 equation:

p^2+8=104

We move all terms to the left:

p^2+8-(104)=0

We add all the numbers together, and all the variables

p^2-96=0

a = 1; b = 0; c = -96;
Δ = b2-4ac
Δ = 02-4·1·(-96)
Δ = 384
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:

$\sqrt{\Delta}=\sqrt{384}=\sqrt{64*6}=\sqrt{64}*\sqrt{6}=8\sqrt{6}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{6}}{2*1}=\frac{0-8\sqrt{6}}{2}
=-\frac{8\sqrt{6}}{2}
=-4\sqrt{6}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{6}}{2*1}=\frac{0+8\sqrt{6}}{2}
=\frac{8\sqrt{6}}{2}
=4\sqrt{6}
$