-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 = 288
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:

$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-12\sqrt{2}}{2*-12}=\frac{24-12\sqrt{2}}{-24}
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+12\sqrt{2}}{2*-12}=\frac{24+12\sqrt{2}}{-24}
$