p^2=98

Solution for p^2=98 equation:

p^2=98

We move all terms to the left:

p^2-(98)=0

a = 1; b = 0; c = -98;
Δ = b2-4ac
Δ = 02-4·1·(-98)
Δ = 392
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:

$\sqrt{\Delta}=\sqrt{392}=\sqrt{196*2}=\sqrt{196}*\sqrt{2}=14\sqrt{2}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-14\sqrt{2}}{2*1}=\frac{0-14\sqrt{2}}{2}
=-\frac{14\sqrt{2}}{2}
=-7\sqrt{2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+14\sqrt{2}}{2*1}=\frac{0+14\sqrt{2}}{2}
=\frac{14\sqrt{2}}{2}
=7\sqrt{2}
$