12+(Px2)=64

Solution for 12+(Px2)=64 equation:

12+(P2)=64

We move all terms to the left:

12+(P2)-(64)=0

We add all the numbers together, and all the variables

P^2-52=0

a = 1; b = 0; c = -52;
Δ = b2-4ac
Δ = 02-4·1·(-52)
Δ = 208
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:

$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{13}}{2*1}=\frac{0-4\sqrt{13}}{2}
=-\frac{4\sqrt{13}}{2}
=-2\sqrt{13}
$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{13}}{2*1}=\frac{0+4\sqrt{13}}{2}
=\frac{4\sqrt{13}}{2}
=2\sqrt{13}
$