The equation $$x^2 - 4x + [x] + 3 = x[x]$$, where $$[x]$$ denotes the greatest integer function, has:

We need to solve $$x^2 - 4x + [x] + 3 = x[x]$$, where $$[x]$$ is the greatest integer function.

Rearrange: $$x^2 - 4x + 3 = x[x] - [x] = [x](x-1)$$

$$(x-1)(x-3) = [x](x-1)$$

Case 1: $$x = 1$$

LHS = 0, RHS = $$[1](0) = 0$$. ✓ $$x = 1$$ is a solution.

Case 2: $$x \neq 1$$

Divide by $$(x-1)$$: $$x - 3 = [x]$$

$$x - [x] = 3$$, i.e., $$\{x\} = 3$$.

But the fractional part $$\{x\} \in [0, 1)$$, so $$\{x\} = 3$$ is impossible.

Therefore, $$x = 1$$ is the unique solution.

Since $$1 \notin (-\infty, 1)$$, the solution is in $$(-\infty, \infty)$$ but not in $$(-\infty, 1)$$.

The correct answer is Option D: a unique solution in $$(-\infty, \infty)$$.