12P=3b^2

Solution for 12P=3b^2 equation:

12=3P^2

We move all terms to the left:

12-(3P^2)=0

a = -3; b = 0; c = +12;
Δ = b2-4ac
Δ = 02-4·(-3)·12
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12}{2*-3}=\frac{-12}{-6}
=+2
$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12}{2*-3}=\frac{12}{-6}
=-2
$