4(j+2)3j=6

Solution for 4(j+2)3j=6 equation:

4(j+2)3j=6

We move all terms to the left:

4(j+2)3j-(6)=0

We multiply parentheses

12j^2+24j-6=0

a = 12; b = 24; c = -6;
Δ = b2-4ac
Δ = 242-4·12·(-6)
Δ = 864
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:

$\sqrt{\Delta}=\sqrt{864}=\sqrt{144*6}=\sqrt{144}*\sqrt{6}=12\sqrt{6}$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-12\sqrt{6}}{2*12}=\frac{-24-12\sqrt{6}}{24}
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+12\sqrt{6}}{2*12}=\frac{-24+12\sqrt{6}}{24}
$