3(2+b)4b=12

Solution for 3(2+b)4b=12 equation:

3(2+b)4b=12

We move all terms to the left:

3(2+b)4b-(12)=0

We add all the numbers together, and all the variables

3(b+2)4b-12=0

We multiply parentheses

12b^2+24b-12=0

a = 12; b = 24; c = -12;
Δ = b2-4ac
Δ = 242-4·12·(-12)
Δ = 1152
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1152}=\sqrt{576*2}=\sqrt{576}*\sqrt{2}=24\sqrt{2}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-24\sqrt{2}}{2*12}=\frac{-24-24\sqrt{2}}{24}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+24\sqrt{2}}{2*12}=\frac{-24+24\sqrt{2}}{24}
$