Step 1: Spot the structure.
We have $f(x) = [x]^2 + 5[x] + 6$, where $[x]$ is the greatest integer function. Notice that $[x]$ is always an integer, so let me call it $n$, where $n$ can be any integer at all.
Step 2: Rewrite using $n$.
Replacing $[x]$ by $n$, the function becomes $g(n) = n^2 + 5n + 6$. Since $x$ ranges over all real numbers, $n = [x]$ ranges over every integer, so $n \in \mathbb{Z}$.
Step 3: Factorise.
The quadratic factors nicely as \[ n^2 + 5n + 6 = (n+2)(n+3). \] So each output is a product of two integers that differ by exactly 1.
Step 4: Make the two factors consecutive integers.
Let me set $k = n + 3$. As $n$ runs through all integers, $k$ also runs through all integers. Then $n + 2 = k - 1$, so \[ (n+2)(n+3) = (k-1)k. \]
Step 5: Describe the set of outputs.
So every value of $f$ is of the form $(k-1)k$ for some integer $k$. That is exactly a product of two consecutive integers, giving values like $0, 2, 6, 12, \dots$ as $k$ varies.
Step 6: Match with the options.
The complete range is therefore $\{\, n \mid n = (k-1)k,\ k \in \mathbb{Z} \,\}$, which is precisely option (C). Nice and clean!
\[ \boxed{\{\, n \mid n = (k-1)k,\ k \in \mathbb{Z} \,\}} \]