30. Let $[x]$ be the greatest integer less than or equal to $x$. Then the number of real solutions to the equation $4 x^{2}-40 \cdot[x]+51=0$ is
(A) 0,
(B) 1;
(C) 23
(D) 33
(E) 4.
30. Since 40 is even, $40[x]-51$ is odd. This implies that $4 x^{2}$ must be an odd integer, let $4 x^{2}=2 k+1$. Thus, $x=\sqrt{2 k+1} / 2$. Substituting it into the original equation, we get $\left[\frac{\sqrt{2 k+1}}{2}\right]=\frac{k+26}{20}$. Therefore, it must be that $k \equiv 14$ (mod 20). On the other hand, $\frac{k+26}{20} \leqslant \frac{\sqrt{2 k+1}}{2}$ $30^{2}$.

Since $x^{2}$ is positive, $k$ is non-negative. Thus, from the first inequality, we get $4 \leqslant k \leqslant 144$, and from the second inequality, we get $k \geq 84$.

Combining these, we have $4 \leqslant k < 24$ or $84 < k \leqslant 144$. In these two intervals, the values of $k$ that satisfy $k \equiv 14 (\bmod 20)$ are $k=14, 94, 114, 134$. The corresponding four solutions are
$$
\begin{array}{l}
\text { the solutions are } \\
x_{1}=\frac{\sqrt{29}}{2}, x_{2}=\frac{\sqrt{189}}{2}, x_{3}=\frac{\sqrt{229}}{2}, \\
x_{4}=\frac{\sqrt{269}}{2} .
\end{array}
$$
(E) True.