10. Let $[m]$ denote the greatest integer not exceeding the real number $m$. Then the sum of all elements in the set $\left\{x \in \mathbf{R} \mid 9 x^{2}-30[x]+20=0\right\}$ is $\qquad$
10. $\sqrt{10}$.
$$
\begin{array}{l}
\text { Given } 9 x^{2}-30[x]+20=0 \\
\Rightarrow 30[x]=9 x^{2}+20>0 \\
\Rightarrow[x]>0 \Rightarrow x>0 . \\
\text { Also }[x] \leqslant x \Rightarrow[x]^{2} \leqslant x^{2}, \text { then } \\
9[x]^{2}-30[x]+20 \leqslant 0 \\
\Rightarrow(3[x]-5)^{2} \leqslant 5 .
\end{array}
$$

Thus, $[x]=1$ or 2.
Substituting these values into the original equation, we get
$$
x_{1}=\frac{\sqrt{10}}{3}, x_{2}=\frac{2 \sqrt{10}}{3} \text {. }
$$

Therefore, the sum of all elements in the set is $x_{1}+x_{2}=\sqrt{10}$.