## Task A-2.2.

Determine all real solutions of the equation

$$
4 x^{2}-20\lfloor x\rfloor+9=0
$$

where $\lfloor x\rfloor$ denotes the greatest integer not greater than $x$.
## Solution.

We have

$$
4 x^{2}-20\lfloor x\rfloor+9 \geqslant 4 x^{2}-20 x+9=(2 x-9)(2 x-1)
$$

so it must be $(2 x-9)(2 x-1) \leqslant 0$, i.e., $\frac{1}{2} \leqslant x \leqslant \frac{9}{2}$. Therefore, $0 \leqslant\lfloor x\rfloor \leqslant 4$.

Also, $x=\frac{1}{2} \sqrt{20\lfloor x\rfloor-9}$.

Substituting $\lfloor x\rfloor=0,1,2,3,4$ in turn. For $\lfloor x\rfloor=0$ there is no solution, and in all other cases we get one value that satisfies the original equation:

$$
x=\frac{1}{2} \sqrt{11}, \quad x=\frac{1}{2} \sqrt{31}, \quad x=\frac{1}{2} \sqrt{51}, \quad x=\frac{1}{2} \sqrt{71}
$$