Determine the unique pair of real numbers $(x,y)$ that satisfy the equation
\[(4 x^2+ 6 x + 4)(4 y^2 - 12 y +   25   )  = 28 .\]
1. Start with the given equation:
   \[
   (4x^2 + 6x + 4)(4y^2 - 12y + 25) = 28
   \]

2. Rewrite the quadratic expressions in a more convenient form by completing the square:
   \[
   4x^2 + 6x + 4 = 4\left(x^2 + \frac{3}{2}x + 1\right)
   \]
   To complete the square inside the parentheses:
   \[
   x^2 + \frac{3}{2}x + 1 = \left(x + \frac{3}{4}\right)^2 - \left(\frac{3}{4}\right)^2 + 1 = \left(x + \frac{3}{4}\right)^2 - \frac{9}{16} + 1 = \left(x + \frac{3}{4}\right)^2 + \frac{7}{16}
   \]
   Therefore:
   \[
   4x^2 + 6x + 4 = 4\left(\left(x + \frac{3}{4}\right)^2 + \frac{7}{16}\right) = 4\left(x + \frac{3}{4}\right)^2 + 4 \cdot \frac{7}{16} = 4\left(x + \frac{3}{4}\right)^2 + \frac{7}{4}
   \]

3. Similarly, for the second quadratic expression:
   \[
   4y^2 - 12y + 25 = 4\left(y^2 - 3y + \frac{25}{4}\right)
   \]
   To complete the square inside the parentheses:
   \[
   y^2 - 3y + \frac{25}{4} = \left(y - \frac{3}{2}\right)^2 - \left(\frac{3}{2}\right)^2 + \frac{25}{4} = \left(y - \frac{3}{2}\right)^2 - \frac{9}{4} + \frac{25}{4} = \left(y - \frac{3}{2}\right)^2 + 4
   \]
   Therefore:
   \[
   4y^2 - 12y + 25 = 4\left(\left(y - \frac{3}{2}\right)^2 + 4\right) = 4\left(y - \frac{3}{2}\right)^2 + 16
   \]

4. Substitute these expressions back into the original equation:
   \[
   \left(4\left(x + \frac{3}{4}\right)^2 + \frac{7}{4}\right)\left(4\left(y - \frac{3}{2}\right)^2 + 16\right) = 28
   \]

5. Notice that both terms inside the parentheses are non-negative and reach their minimum values when the squared terms are zero:
   \[
   \left(4\left(x + \frac{3}{4}\right)^2 + \frac{7}{4}\right) \geq \frac{7}{4}
   \]
   \[
   \left(4\left(y - \frac{3}{2}\right)^2 + 16\right) \geq 16
   \]

6. The product of these minimum values is:
   \[
   \frac{7}{4} \cdot 16 = 28
   \]

7. Equality holds if and only if both squared terms are zero:
   \[
   4\left(x + \frac{3}{4}\right)^2 = 0 \quad \text{and} \quad 4\left(y - \frac{3}{2}\right)^2 = 0
   \]
   Solving these equations:
   \[
   x + \frac{3}{4} = 0 \implies x = -\frac{3}{4}
   \]
   \[
   y - \frac{3}{2} = 0 \implies y = \frac{3}{2}
   \]

8. Therefore, the unique solution is:
   \[
   (x, y) = \left(-\frac{3}{4}, \frac{3}{2}\right)
   \]

The final answer is \( \boxed{ \left(-\frac{3}{4}, \frac{3}{2}\right) } \).