Example 4 Given that $a$, $b$, and $2$ are the side lengths of a triangle, and $a$, $b$ are the roots of the equation
$$
\left(3 x^{2}-4 x-1\right)\left(3 x^{2}-4 x-5\right)=12
$$
Then the perimeter of the triangle can only be ( ).
(A) $\frac{10}{3}$ or $\frac{8}{3}$
(B) $\frac{14}{3}$ or $\frac{10}{3}$
(C) $\frac{16}{3}$ or $\frac{14}{3}$
(D) $\frac{16}{3}$ or $\frac{20}{3}$
Let $3 x^{2}-4 x-1=y$.
Then the original equation becomes $y(y-4)=12$.
Solving this equation yields $y=-2$ or 6.
Thus, $3 x^{2}-4 x-1=-2$ or $3 x^{2}-4 x-1=6$.
Solving these gives $x_{1}=1, x_{2}=\frac{1}{3}, x_{3}=\frac{7}{3}, x_{4}=-1$.
If $a \neq b$, from $a+b>2,|a-b|2$, we know $a=b=\frac{7}{3}$.
Therefore, the perimeter of the triangle is $\frac{16}{3}$ or $\frac{20}{3}$. The answer is D.