Find all the real roots of
\[(12x - 1)(6x - 1)(4x - 1)(3x - 1) = 5.\]Enter all the real roots, separated by commas.
We can expand using the pairs $(3x - 1)(12x - 1) = 36x^2 - 15x + 1$ and $(6x - 1)(4x - 1) = 24x^2 - 10x + 1,$ so
\[(36x^2 - 15x + 1)(24x^2 - 10x + 1) = 5.\]Let $y = 12x^2 - 5x.$  Then
\[(3y + 1)(2y + 1) = 5.\]This simplifies to $6y^2 + 5y - 4 = 0,$ which factors as $(2y - 1)(3y + 4) = 0.$  Hence, $y = \frac{1}{2}$ or $y = -\frac{4}{3}.$

If $12x^2 - 5x = \frac{1}{2},$ then $24x^2 - 10x - 1 = 0,$ which factors as
\[(2x - 1)(12x + 1) = 0.\]Hence, $x = \frac{1}{2}$ or $x = -\frac{1}{12}.$

If $12x^2 - 5x = -\frac{4}{3},$ then
\[36x^2 - 15x + 4 = 0,\]which has no real solutions.

Therefore, the real roots are $\boxed{\frac{1}{2}, -\frac{1}{12}}.$