4p^2-4=68

Solution for 4p^2-4=68 equation:

4p^2-4=68

We move all terms to the left:

4p^2-4-(68)=0

We add all the numbers together, and all the variables

4p^2-72=0

a = 4; b = 0; c = -72;
Δ = b2-4ac
Δ = 02-4·4·(-72)
Δ = 1152
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:

$\sqrt{\Delta}=\sqrt{1152}=\sqrt{576*2}=\sqrt{576}*\sqrt{2}=24\sqrt{2}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24\sqrt{2}}{2*4}=\frac{0-24\sqrt{2}}{8}
=-\frac{24\sqrt{2}}{8}
=-3\sqrt{2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24\sqrt{2}}{2*4}=\frac{0+24\sqrt{2}}{8}
=\frac{24\sqrt{2}}{8}
=3\sqrt{2}
$