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

Clearly, that $x=\boxed{0}$ is one solution. Cancelling $x$ on both sides and rearrange gives $$\begin{align} x^3 - 6x^2 + 9x -4 = 0\end{align}$$

Note that the sum of all the coefficents equals $0$, therefore $x=\boxed{1}$ is one root. From there we know that $$x^3 - 6x^2 + 9x -4=(x-1)^2(x-4)$$

Therefore all the roots are $x=\boxed{0, 1, 4}$.