5. The number of real numbers $x$ that satisfy the equation $36 x^{2}-140[x]-21=0$ (where $[x]$ denotes the greatest integer less than or equal to $x$) is ( ).
(A) 2
(B) 3
(C) 4
(D) 5
5.C.

Let $x=k+\alpha(0 \leqslant \alpha < 1, k \geqslant 0)$, i.e., the solution of the equation is a positive number.
According to equation (1), we have
$$
\begin{array}{l}
21=36(k+\alpha)^{2}-140 k>36 k^{2}-140 k \\
=36 k\left(k-\frac{35}{9}\right)=36\left(k-\frac{35}{18}\right)^{2}-36\left(\frac{35}{18}\right)^{2} .
\end{array}
$$

Therefore, when $k>\frac{35}{18}$, the right side of the above inequality is an increasing function of $k$.
If $k \geqslant 5$, then
the right side of equation (2) $>36 \times 5\left(5-\frac{35}{9}\right)=200$,
which is a contradiction.
Thus, $0 \leqslant k \leqslant 4$.
From equation (1), we get
$$
x=k+\alpha=\frac{\sqrt{140 k+21}}{6} \text {. }
$$

By taking $k=0,1,2,3,4$, we obtain
$$
x=\frac{\sqrt{21}}{6}, \frac{\sqrt{161}}{6}, \frac{\sqrt{301}}{6}, \frac{7}{2}, \frac{\sqrt{581}}{6} \text {. }
$$

Upon verification, $x=\frac{\sqrt{161}}{6}$ does not satisfy the problem's conditions.
Therefore, there are 4 real numbers that satisfy the original equation.