Problems of Ukrainian Authors

Let's build the graph of the equation, namely the function $y = x^2 - 3x[x] + 2x = a$. It is easy to understand that we are interested only when $x > 0$.
$$\begin{align*} x \in [0, 1) &\Rightarrow [x] = 0 \Rightarrow y = x^2 + 2x \\ x \in [1, 2) &\Rightarrow [x] = 1 \Rightarrow y = x^2 - x \\ x \in [2, 3) &\Rightarrow [x] = 2 \Rightarrow y = x^2 - 4x \\ x \in [3, 4) &\Rightarrow [x] = 3 \Rightarrow y = x^2 - 7x \end{align*}$$
While $x \ge 4$, the apex of the parabola $y = x^2 + x(2 - 3[x])$ is situated at the point $x_b = \frac{3|x| - 2}{2} \ge [x] + 1$, which is equivalent to $[x] \ge 4$. From this, it follows that the function on every interval like $[n, n + 1)$ under the condition $n \ge 4$ is downward. In addition, it is downward when $x \ge 4$, which is shown by the following transformations:
$$x \in [n - 1, n) \Rightarrow [x] = n - 1 \Rightarrow y_{n-1} = x^2 + x(5 - 3n) \\ \text{and } y_{n-1}^{\min} > y_{n-1}(n) = n^2 + n(5 - 3n);$$
$$x \in [n, n + 1) \Rightarrow [x] = n \Rightarrow y_n = x^2 + x(2 - 3n) \text{ and } y_n^{\max} = y_n(n) = n^2 + n(2 - 3n) < y_{n-1}^{\min}.$$
Thus, exactly two positive solutions of the function can exist only when $x \le 4$, and here it is easy to portray the study graph function. For easy perception, the scale is changed. So we can see exactly two positive solutions under the condition $0 < a < 2$, also on the interval $-\frac{49}{4} < a < -12$.