p2=125

Solution for p2=125 equation:

p2=125

We move all terms to the left:

p2-(125)=0

We add all the numbers together, and all the variables

p^2-125=0

a = 1; b = 0; c = -125;
Δ = b2-4ac
Δ = 02-4·1·(-125)
Δ = 500
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:

$\sqrt{\Delta}=\sqrt{500}=\sqrt{100*5}=\sqrt{100}*\sqrt{5}=10\sqrt{5}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{5}}{2*1}=\frac{0-10\sqrt{5}}{2}
=-\frac{10\sqrt{5}}{2}
=-5\sqrt{5}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{5}}{2*1}=\frac{0+10\sqrt{5}}{2}
=\frac{10\sqrt{5}}{2}
=5\sqrt{5}
$