To solve the equation $\lfloor x^2 - 2x \rfloor + 2\lfloor x \rfloor = \lfloor x \rfloor^2$, we proceed with the following steps:

### Step 1: Understanding the Floor Function

The floor function, $\lfloor y \rfloor$, represents the greatest integer less than or equal to $y$. This means that for any real number $x$, $\lfloor x \rfloor \leq x < \lfloor x \rfloor + 1$.

### Step 2: Set $x = n + \theta$

Consider $x = n + \theta$ where $n = \lfloor x \rfloor$ and $0 \leq \theta < 1$. Substituting this into the main equation gives:

$$\lfloor (n+\theta)^2 - 2(n+\theta) \rfloor + 2n = n^2.$$

This simplifies to:

$$\lfloor n^2 + 2n\theta + \theta^2 - 2n - 2\theta \rfloor + 2n = n^2.$$

Hence, we have:

$$\lfloor \theta^2 + \theta(2n - 2) \rfloor = 0.$$

This equation implies that:

$$0 \leq \theta^2 + \theta(2n - 2) < 1.$$

### Step 3: Solve the Quadratic Inequality

The inequality becomes:

$$\theta^2 + \theta(2n - 2) < 1.$$

Completing the square, we have:

$$(\theta + (n-1))^2 < (n-1)^2 + 1.$$

For $0 \leq \theta < 1$, $\theta + (n-1) < \sqrt{(n-1)^2 + 1}$. Therefore, this restricts:

$$n - 1 < \theta + (n-1) < \sqrt{(n-1)^2 + 1}.$$

This inequality in terms of $\theta$ results in:

$$n - 1 < x < \sqrt{(n-1)^2 + 1} + 1.$$

### Conclusion

The above range, $n - 1 < x < \sqrt{(n-1)^2 + 1} + 1$, corresponds to the solution intervals $(n, \sqrt{(n-1)^2 + 1} + 1)$ for each integer $n \geq 1$. Additionally, since $n$ is an integer, $x$ can also be any integer, hence $x \in \mathbb{Z}$.

Thus, the complete set of solutions is:

$$\boxed{x \in \mathbb{Z} \cup \bigcup_{n = 1} ^{\infty} (n, \sqrt{(n-1)^2 + 1} + 1)}.$$