2. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Then the number of all real solutions to the equation $[2 x]+[3 x]=8 x-6$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4
2. C.

When $x$ is an integer,
$$
2 x+3 x=8 x-6 \Rightarrow x=2 \text {. }
$$

When $x$ is not an integer, let $x=t+r(t$ be an integer, $0<r<1)$.
(1) If $0<r<\frac{1}{3}$, then
$$
\begin{array}{l}
2 t+3 t=8(t+r)-6 \\
\Rightarrow t=2-\frac{8}{3} r \\
\Rightarrow \frac{10}{9}<t=2-\frac{8}{3} r<2
\end{array}
$$
$\Rightarrow t$ has no integer solution;
(2) If $\frac{1}{3} \leqslant r<\frac{1}{2}$, then
$$
\begin{array}{l}
2 t+3 t+1=8(t+r)-6 \\
\Rightarrow t=\frac{7}{3}-\frac{8}{3} r \Rightarrow 1<t \leqslant \frac{13}{9}
\end{array}
$$
$\Rightarrow t$ has no integer solution;
(3) If $\frac{1}{2} \leqslant r<\frac{2}{3}$, then
$$
\begin{array}{l}
2 t+1+3 t+1=8(t+r)-6 \\
\Rightarrow t=\frac{8}{3}-\frac{8}{3} r \Rightarrow \frac{8}{9}<t \leqslant \frac{4}{3} \\
\Rightarrow t=1, r=\frac{5}{8} \Rightarrow x=\frac{13}{8}
\end{array}
$$
(4) If $\frac{2}{3} \leqslant r<1$, then
$$
\begin{array}{l}
2 t+1+3 t+2=8(t+r)-6 \\
\Rightarrow t=3-\frac{8}{3} r \Rightarrow \frac{1}{3}<t \leqslant \frac{11}{9} \\
\Rightarrow t=1, r=\frac{3}{4} \Rightarrow x=\frac{7}{4} .
\end{array}
$$

In summary, there are 3 real solutions that meet the conditions.