3j^2+12j+6=0

Solution for 3j^2+12j+6=0 equation:

3j^2+12j+6=0

a = 3; b = 12; c = +6;
Δ = b2-4ac
Δ = 122-4·3·6
Δ = 72
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-6\sqrt{2}}{2*3}=\frac{-12-6\sqrt{2}}{6}
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+6\sqrt{2}}{2*3}=\frac{-12+6\sqrt{2}}{6}
$