Given the equation \[ y^4 \plus{} 4y^2x \minus{} 11y^2 \plus{} 4xy \minus{} 8y \plus{} 8x^2 \minus{} 40x \plus{} 52 \equal{} 0,\] find all real solutions.
Given the equation:
\[ y^4 + 4y^2x - 11y^2 + 4xy - 8y + 8x^2 - 40x + 52 = 0, \]

we need to find all real solutions \((x, y)\).

1. **Rearrange the equation:**
   \[ 8x^2 + (4y^2 + 4y - 40)x + (y^4 - 11y^2 - 8y + 52) = 0. \]

2. **Complete the square:**
   Notice that we can rewrite the equation in a form that allows us to complete the square:
   \[ (4x)^2 + 2(4x)(y^2 + y - 10) + (y^2 + y - 10)^2 + 2(y^4 - 11y^2 - 8y + 52) - (y^2 + y - 10)^2 = 0. \]

3. **Simplify the equation:**
   \[ (4x + y^2 + y - 10)^2 + (y^2 - y - 2)^2 = 0. \]

4. **Analyze the equation:**
   Since both terms \((4x + y^2 + y - 10)^2\) and \((y^2 - y - 2)^2\) are squares and squares of real numbers are non-negative, the only way their sum can be zero is if both terms are zero:
   \[ 4x + y^2 + y - 10 = 0, \]
   \[ y^2 - y - 2 = 0. \]

5. **Solve the second equation:**
   \[ y^2 - y - 2 = 0. \]
   Factorize:
   \[ (y - 2)(y + 1) = 0. \]
   Therefore, the solutions for \(y\) are:
   \[ y = 2 \quad \text{or} \quad y = -1. \]

6. **Substitute \(y = 2\) into the first equation:**
   \[ 4x + (2)^2 + 2 - 10 = 0, \]
   \[ 4x + 4 + 2 - 10 = 0, \]
   \[ 4x - 4 = 0, \]
   \[ 4x = 4, \]
   \[ x = 1. \]

7. **Substitute \(y = -1\) into the first equation:**
   \[ 4x + (-1)^2 + (-1) - 10 = 0, \]
   \[ 4x + 1 - 1 - 10 = 0, \]
   \[ 4x - 10 = 0, \]
   \[ 4x = 10, \]
   \[ x = 2.5. \]

Therefore, the real solutions are:
\[ (x, y) = (1, 2) \quad \text{and} \quad (2.5, -1). \]

The final answer is \(\boxed{(1, 2) \text{ and } (2.5, -1)}\).