Let [t] denote the greatest integer $$\leq$$ t. Then the equation in x, $$[x]^2 + 2[x+2] - 7 = 0$$ has:

1. Simplify the Equation

We are given the equation:

$$[x]^2 + 2[x + 2] - 7 = 0$$

Using the property of the greatest integer function $$[x + n] = [x] + n$$ (where $$n$$ is an integer), we can rewrite $$[x+2]$$ as $$[x] + 2$$:

$$[x]^2 + 2([x] + 2) - 7 = 0$$

Expand and simplify:

$$[x]^2 + 2[x] + 4 - 7 = 0$$

$$[x]^2 + 2[x] - 3 = 0$$

2. Solve the Quadratic Equation

Let $$y = [x]$$. The equation becomes a standard quadratic:

$$y^2 + 2y - 3 = 0$$

Factoring the quadratic:

$$(y + 3)(y - 1) = 0$$

This gives us two possible values for $$[x]$$:

1. $$[x] = 1$$
2. $$[x] = -3$$

3. Analyze the Solutions for $$x$$

The greatest integer function $$[x] = k$$ means that $$x$$ can be any real number in the interval $$k \le x < k + 1$$.

• Case 1: If $$[x] = 1$$, then:$$1 \le x < 2$$

  This interval contains infinitely many real numbers.

• Case 2: If $$[x] = -3$$, then:$$-3 \le x < -2$$

  This interval also contains infinitely many real numbers.

Final Conclusion

Since $$x$$ can be any value within the intervals $$[1, 2)$$ or $$[-3, -2)$$, there are an infinitely many solutions for $$x$$.

The correct option is D.