Let us solve the equation

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

where $[t]$ denotes the greatest integer less than or equal to $t$.

1. Use the property of greatest integer function

Since $2$ is an integer,

$[x+2] = [x] + 2.$

So the given equation becomes

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

Simplify:

$[x]^2 + 2[x] + 4 - 7 = 0$ $[x]^2 + 2[x] - 3 = 0.$

2. Substitute $n = [x]$

Let

$n = [x],$

where $n$ is an integer. Then we get

$n^2 + 2n - 3 = 0.$

Factorizing,

$n^2 + 2n - 3 = (n+3)(n-1) = 0.$

Hence,

$n = 1 \quad \text{or} \quad n = -3.$

So,

$[x] = 1 \quad \text{or} \quad [x] = -3.$

3. Convert back to ranges of $x$

Case 1: $[x] = 1$

This means

$1 \le x < 2.$

Case 2: $[x] = -3$

This means

$-3 \le x < -2.$

Thus the solution set is

$x \in [-3,-2) \cup [1,2).$

4. Count the number of solutions

Each of the intervals $[-3,-2)$ and $[1,2)$ contains infinitely many real numbers. Therefore, the equation has infinitely many solutions.

5. Check the options

• A: no integral solution — false, because $x=-3,-2.5,1,1.7$ etc. are solutions, and in particular $x=-3,1$ are integral solutions.
• B: exactly two solutions — false.
• C: exactly four integral solutions — false. Integral solutions are only $x=-3$ and $x=1$, i.e. two integral solutions.
• D: infinitely many solutions — true.

Final Answer

The correct option is

$\boxed{D}$

because the solution set is

$\boxed{[-3,-2) \cup [1,2)}.$