p^2=62

Solution for p^2=62 equation:

p^2=62

We move all terms to the left:

p^2-(62)=0

a = 1; b = 0; c = -62;
Δ = b2-4ac
Δ = 02-4·1·(-62)
Δ = 248
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:

$\sqrt{\Delta}=\sqrt{248}=\sqrt{4*62}=\sqrt{4}*\sqrt{62}=2\sqrt{62}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{62}}{2*1}=\frac{0-2\sqrt{62}}{2}
=-\frac{2\sqrt{62}}{2}
=-\sqrt{62}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{62}}{2*1}=\frac{0+2\sqrt{62}}{2}
=\frac{2\sqrt{62}}{2}
=\sqrt{62}
$