3p^2+12p=6

Solution for 3p^2+12p=6 equation:

3p^2+12p=6

We move all terms to the left:

3p^2+12p-(6)=0

a = 3; b = 12; c = -6;
Δ = b2-4ac
Δ = 122-4·3·(-6)
Δ = 216
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:

$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-6\sqrt{6}}{2*3}=\frac{-12-6\sqrt{6}}{6}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+6\sqrt{6}}{2*3}=\frac{-12+6\sqrt{6}}{6}
$