Let's determine all solutions of the following equation, where $[x]$ is the greatest integer not exceeding $x$ (for example, $[-3.7]=-4$).

$$
[x]-\sqrt{\frac{[x]}{x-0.5}}-\frac{6}{x-0.5}=0
$$
It is obvious that $x \neq 0.5$, because otherwise the fractions in the equation would not be defined. Multiply the equation by the expression $(x-0.5)$

$$
[x](x-0.5)-(x-0.5) \sqrt{\frac{[x]}{x-0.5}}-6=0
$$

Case I. If $x-0.5>0$, then $x-0.5=\sqrt{(x-0.5)^{2}}$, and thus

$$
[x](x-0.5)-\sqrt{[x](x-0.5)}-6=0
$$

This is a quadratic equation in terms of $\sqrt{[x](x-0.5)}$, whose roots are 3 and -2. The value of the square root expression cannot be negative, so

$$
\sqrt{[x](x-0.5)}=3, \quad \text { from which } \quad[x](x-0.5)=9 .
$$

It is clear that as the value of $x$ increases, the value of $[x](x-0.5)$ also increases, because both factors are increasing and non-negative. This implies that the solution to the equation $[x](x-0.5)=9$ should be sought in the interval $3<x<4$. Here $[x]=3$, from which $x=3.5$ follows. This indeed satisfies the original equation.

Case II. If $x-0.5<0$, then $x-0.5=-\sqrt{(x-0.5)^{2}}$, and thus we arrive at the following equation:

$$
[x](x-0.5)+\sqrt{[x](x-0.5)}-6=0
$$

Solving similarly to the first case, now $\sqrt{[x](x-0.5)}=2$, and from this the equation $[x](x-0.5)=4$ follows. As the variable $x$ increases, now the product decreases, because both factors are increasing and non-positive. It is immediately apparent that the solution now falls between -2 and -1, where $[x]=-2$. Thus $x=-1.5$, which indeed satisfies the original equation. In summary, the equation has two solutions, and these are $x=-1.5$ and $x=3.5$.

Janos Filus (Kecskemét, Katona J. Gymnasium, 2nd grade)