6. The solution to the equation $\sqrt{3+\sqrt{3+\sqrt{3+x}}}=x$ is
$\qquad$ .
6. $\frac{1+\sqrt{13}}{2}$.

Obviously, $x>0$.
First, prove: If $x>\sqrt{3+x}$, then the equation has no solution.
In fact, in this case, we have
$$
\begin{array}{l}
\sqrt{3+\sqrt{3+\sqrt{3+x}}}<\sqrt{3+\sqrt{3+x}} \\
<\sqrt{3+x}<x .
\end{array}
$$

Similarly, when $x<\sqrt{3+x}$, the equation also has no solution.
Therefore, $x=\sqrt{3+x}$, i.e., $x=\frac{1+\sqrt{13}}{2}$.