2p^2+34=64

Solution for 2p^2+34=64 equation:

2p^2+34=64

We move all terms to the left:

2p^2+34-(64)=0

We add all the numbers together, and all the variables

2p^2-30=0

a = 2; b = 0; c = -30;
Δ = b2-4ac
Δ = 02-4·2·(-30)
Δ = 240
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:

$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{15}}{2*2}=\frac{0-4\sqrt{15}}{4}
=-\frac{4\sqrt{15}}{4}
=-\sqrt{15}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{15}}{2*2}=\frac{0+4\sqrt{15}}{4}
=\frac{4\sqrt{15}}{4}
=\sqrt{15}
$