109 Let $[x]$ denote the greatest integer not greater than $x$. Then the solution set of the equation $4 x^{2}-40[x]+51=0$ is
$\qquad$
$109\left\{\frac{1}{2} \sqrt{29}, \frac{1}{2} \sqrt{189}, \frac{1}{2} \sqrt{229}, \frac{1}{2} \sqrt{269}\right\}$. Clearly $x>0$, let $[x]=t$, then
$$
4 x^{2}-40 t+51=0,
$$

so
$$
x=\sqrt{\frac{40 t-51}{4}},
$$

since
$$
[x] \leqslant x<[x]+1,
$$

we have
$$
t \leqslant \sqrt{\frac{40 t-51}{4}}<t+1,
$$

solving this yields $\frac{3}{2}<t<\frac{5}{2}$ or $\frac{11}{2}<t<\frac{17}{2}$. Since $t \in \mathbf{N}$, we have $t=2,6,7,8$, further we get
$$
x_{1}=\frac{1}{2} \sqrt{29}, x_{2}=\frac{1}{2} \sqrt{189}, x_{3}=\frac{1}{2} \sqrt{229}, x_{4}=\frac{1}{2} \sqrt{269} .
$$