14. $[x]$ represents the greatest integer not exceeding the real number $x$. The equation $[2 x]+[3 x]=9 x-\frac{7}{4}$ has all real solutions as $\qquad$
14. $-\frac{1}{36}, \frac{7}{36}$.

From $2 x-1<[2 x] \leqslant 2 x, 3 x-1<[3 x] \leqslant 3 x$, then $5 x-2<9 x-\frac{7}{4} \leqslant 5 x \Rightarrow-\frac{1}{16}<x \leqslant \frac{7}{16}$. Therefore, $-\frac{37}{16}<9 x-\frac{7}{4} \leqslant \frac{35}{16}$.
Since $9 x-\frac{7}{4}$ is an integer, we have,
$$
9 x-\frac{7}{4}=-2,-1,0,1,2 \text {. }
$$

Solving for $x$ gives $x=-\frac{1}{36}, \frac{1}{12}, \frac{7}{36}, \frac{11}{36}, \frac{5}{12}$.
Verification yields $x=-\frac{1}{36}, \frac{7}{36}$.