13. Solve the equation $4 x^{2}-40[x]+51=0$.
13. Solution: Let $[x]=n,\{x\}=\alpha$, then $x=n+\alpha, \alpha \in[0,1)$, so the original equation becomes $4 x^{2}-40 x+51=$ $-40 \alpha$. Since $\alpha \in[0,1)$, we have $-40<4 x^{2}-40 x+51 \leqslant 0$. Solving this, we get $\frac{3}{2} \leqslant x<\frac{7}{2}$ or $\frac{13}{2}<x \leqslant \frac{17}{2}$.
(1) When $\frac{3}{2} \leqslant x<2$, $n=1,4 x^{2}-40+51=0$, no real solutions;
(2) When $2 \leqslant x<3$, $[x]=2,4 x^{2}-80+51=0, x=\frac{\sqrt{29}}{2}$ (discard $-\frac{\sqrt{29}}{2}$);
(3) When $3 \leqslant x<\frac{7}{2}$, $n=3,4 x^{2}-120+51=0, x= \pm \frac{\sqrt{69}}{2}$ (discard);
(4) When $\frac{13}{2} \leqslant x<7$, $n=6,4 x^{2}-240+51=0, x=\frac{\sqrt{189}}{2}$ (discard $-\frac{\sqrt{189}}{2}$);
(5) When $7 \leqslant x<8$, $n=7,4 x^{2}-280+51=0, x=\frac{\sqrt{229}}{2}$ (discard $-\frac{\sqrt{229}}{2}$);
(6) When $8 \leqslant x<\frac{17}{2}$, $n=8,4 x^{2}-320+51=0, x=\frac{\sqrt{269}}{2}$ (discard $-\frac{\sqrt{269}}{2}$).
Therefore, the original equation has 4 solutions: $x_{1}=\frac{\sqrt{29}}{2}, x_{2}=\frac{\sqrt{189}}{2}, x_{3}=\frac{\sqrt{229}}{2}, x_{4}=\frac{\sqrt{269}}{2}$.