Find all real numbers $ x$ such that:
$ [x^2\plus{}2x]\equal{}{[x]}^2\plus{}2[x]$

(Here $ [x]$ denotes the largest integer not exceeding $ x$.)
1. Given the equation:
   \[
   [x^2 + 2x] = [x]^2 + 2[x]
   \]
   where \([x]\) denotes the greatest integer function (floor function).

2. Let \( n = [x] \). Then \( n \leq x < n+1 \).

3. Substitute \( n \) into the equation:
   \[
   [x^2 + 2x] = n^2 + 2n
   \]

4. Consider the expression \( x^2 + 2x \):
   \[
   n^2 \leq x^2 < (n+1)^2
   \]
   \[
   n^2 + 2n \leq x^2 + 2x < (n+1)^2 + 2(n+1)
   \]

5. Since \( n \leq x < n+1 \), we have:
   \[
   n^2 + 2n \leq x^2 + 2x < n^2 + 4n + 3
   \]

6. The floor function \([x^2 + 2x]\) must be an integer, so:
   \[
   n^2 + 2n \leq x^2 + 2x < n^2 + 2n + 1
   \]

7. This implies:
   \[
   n^2 + 2n \leq x^2 + 2x < n^2 + 2n + 1
   \]

8. Therefore, \( x^2 + 2x \) must be in the interval \([n^2 + 2n, n^2 + 2n + 1)\).

9. Completing the square for \( x^2 + 2x \):
   \[
   x^2 + 2x = (x+1)^2 - 1
   \]

10. Let \( y = x + 1 \). Then:
    \[
    (x+1)^2 - 1 = y^2 - 1
    \]

11. Substitute \( y \) into the interval:
    \[
    n^2 + 2n \leq y^2 - 1 < n^2 + 2n + 1
    \]

12. Adding 1 to all parts of the inequality:
    \[
    n^2 + 2n + 1 \leq y^2 < n^2 + 2n + 2
    \]

13. Since \( y^2 \) is an integer, it must be equal to \( n^2 + 2n + 1 \):
    \[
    y^2 = n^2 + 2n + 1
    \]

14. Therefore:
    \[
    (x+1)^2 = n^2 + 2n + 1
    \]

15. Solving for \( x \):
    \[
    x + 1 = n + 1 \quad \text{or} \quad x + 1 = -(n + 1)
    \]

16. Thus:
    \[
    x = n \quad \text{or} \quad x = -n - 2
    \]

17. Since \( n = [x] \), \( x = n \) is valid. For \( x = -n - 2 \), it must satisfy \( n \leq -n - 2 < n+1 \), which is not possible for any integer \( n \).

18. Therefore, the only solution is:
    \[
    x = n
    \]

The final answer is \( \boxed{ x = n } \) where \( n \) is an integer.