Let $a$ be a fixed real number. Consider the equation
$$
(x+2)^{2}(x+7)^{2}+a=0, x \in R
$$
where $R$ is the set of real numbers. For what values of $a$, will the equ have exactly one double-root?
1. Define the function \( f(x) = (x+2)^2 (x+7)^2 + a \).
2. Compute the first derivative of \( f(x) \):
   \[
   f'(x) = \frac{d}{dx} \left[ (x+2)^2 (x+7)^2 + a \right] = 2(x+2)(x+7)(2x+9)
   \]
3. Find the critical points by setting \( f'(x) = 0 \):
   \[
   2(x+2)(x+7)(2x+9) = 0
   \]
   This gives us the critical points \( x = -2 \), \( x = -7 \), and \( x = -\frac{9}{2} \).

4. Determine the nature of these critical points by evaluating the second derivative \( f''(x) \):
   \[
   f''(x) = \frac{d}{dx} \left[ 2(x+2)(x+7)(2x+9) \right]
   \]
   Using the product rule:
   \[
   f''(x) = 2 \left[ (x+7)(2x+9) + (x+2)(2x+9) + (x+2)(x+7) \cdot 2 \right]
   \]
   Simplifying:
   \[
   f''(x) = 2 \left[ 2(x+7) + 2(x+2) + 2(x+2)(x+7) \right]
   \]
   \[
   f''(x) = 2 \left[ 2x + 14 + 2x + 4 + 2(x^2 + 9x + 14) \right]
   \]
   \[
   f''(x) = 2 \left[ 4x + 18 + 2x^2 + 18x + 28 \right]
   \]
   \[
   f''(x) = 2 \left[ 2x^2 + 22x + 46 \right]
   \]
   \[
   f''(x) = 4x^2 + 44x + 92
   \]

5. Evaluate \( f''(x) \) at the critical points:
   - At \( x = -2 \):
     \[
     f''(-2) = 4(-2)^2 + 44(-2) + 92 = 16 - 88 + 92 = 20 \quad (\text{local minimum})
     \]
   - At \( x = -7 \):
     \[
     f''(-7) = 4(-7)^2 + 44(-7) + 92 = 196 - 308 + 92 = -20 \quad (\text{local minimum})
     \]
   - At \( x = -\frac{9}{2} \):
     \[
     f''\left(-\frac{9}{2}\right) = 4\left(-\frac{9}{2}\right)^2 + 44\left(-\frac{9}{2}\right) + 92 = 81 - 198 + 92 = -25 \quad (\text{local maximum})
     \]

6. Since \( f(x) \) has local minima at \( x = -2 \) and \( x = -7 \), and a local maximum at \( x = -\frac{9}{2} \), we need \( f\left(-\frac{9}{2}\right) = 0 \) for the function to have exactly one double root.

7. Calculate \( f\left(-\frac{9}{2}\right) \):
   \[
   f\left(-\frac{9}{2}\right) = \left(-\frac{9}{2} + 2\right)^2 \left(-\frac{9}{2} + 7\right)^2 + a
   \]
   \[
   = \left(-\frac{5}{2}\right)^2 \left(\frac{5}{2}\right)^2 + a
   \]
   \[
   = \left(\frac{25}{4}\right) \left(\frac{25}{4}\right) + a
   \]
   \[
   = \frac{625}{16} + a
   \]
   Set this equal to 0:
   \[
   \frac{625}{16} + a = 0
   \]
   \[
   a = -\frac{625}{16}
   \]
   \[
   a = -39.0625
   \]

The final answer is \( \boxed{ a = -39.0625 } \)