The given equation is equivalent to $$\begin{align}\sqrt{3x^2 + 1} - 2\sqrt{x} &= x-1 \\ 3x^2 + 4x + 1 -4\sqrt{x(3x^2+1)} & = x^2 - 2x + 1 \\ 2\sqrt{x(3x^2 + 1)} &= x^2 + 3x \\12x^3 + 4x &= x^4 +6x^3 + 9x^2 \end{align}$$

It is now clear that $x=0$ is one root candidate. Cancelling $x$ from both sides and re-arranging give: $$\begin{align}x^3 -6x^2 + 9x -4=0\\(x-1)^2(x-4)&=0\end{align}$$

Therefore, all the candidates are $x=0, 1, 4$. Setting each of them back to the original equation finds only $x=\boxed{1, 4}$ are valid roots.