14.11 $[a]$ represents the greatest integer not greater than the number $a$, for example $[\sqrt{2}]=1$, $[-\sqrt{2}]=-2$. Then, what is the sum of all roots of the equation
$$
[3 x+1]=2 x-\frac{1}{2}
$$

(China Junior High School Mathematics League, 1987)
[Solution] Let $[3 x+1]=t$, then $t$ is an integer, and
$$
0 \leqslant(3 x+1)-t<1 \text {. }
$$

Thus, the original equation can be transformed into
$$
\begin{array}{l}
t=2 x-\frac{1}{2} \\
x=\frac{1}{2} t+\frac{1}{4}
\end{array}
$$

Substituting into inequality (1) yields
$$
\begin{array}{l} 
0 \leqslant \frac{3}{2} t+\frac{3}{4}+1-t<1 \\
-\frac{7}{2} \leqslant t \leqslant-\frac{3}{2} .
\end{array}
$$

The integer $t$ that satisfies (2) is
$$
t=-2 \text { or } t=-3 \text {. }
$$

When $t=-2$, $x_{1}=-\frac{3}{4}$.
When $t=-3$, $x_{2}=-\frac{5}{4}$.
Therefore, $x_{1}+x_{2}=\left(-\frac{3}{4}\right)+\left(-\frac{5}{4}\right)=-2$.