Find all real numbers $ a,b,c,d$ such that \[ \left\{\begin{array}{cc}a \plus{} b \plus{} c \plus{} d \equal{} 20, \\
ab \plus{} ac \plus{} ad \plus{} bc \plus{} bd \plus{} cd \equal{} 150. \end{array} \right.\]
1. We start with the given system of equations:
   \[
   \begin{cases}
   a + b + c + d = 20 \\
   ab + ac + ad + bc + bd + cd = 150
   \end{cases}
   \]

2. We know from algebraic identities that:
   \[
   (a + b + c + d)^2 = a^2 + b^2 + c^2 + d^2 + 2(ab + ac + ad + bc + bd + cd)
   \]

3. Substituting the given values into the identity:
   \[
   20^2 = a^2 + b^2 + c^2 + d^2 + 2 \cdot 150
   \]

4. Simplifying the equation:
   \[
   400 = a^2 + b^2 + c^2 + d^2 + 300
   \]

5. Solving for \(a^2 + b^2 + c^2 + d^2\):
   \[
   a^2 + b^2 + c^2 + d^2 = 400 - 300 = 100
   \]

6. Using the Cauchy-Schwarz inequality:
   \[
   (a^2 + b^2 + c^2 + d^2)(1^2 + 1^2 + 1^2 + 1^2) \geq (a + b + c + d)^2
   \]

7. Substituting the known values:
   \[
   4(a^2 + b^2 + c^2 + d^2) \geq (a + b + c + d)^2
   \]

8. Simplifying with the given sums:
   \[
   4 \cdot 100 \geq 20^2
   \]

9. This simplifies to:
   \[
   400 \geq 400
   \]

10. Since equality holds in the Cauchy-Schwarz inequality, it implies that \(a = b = c = d\).

11. Given \(a + b + c + d = 20\) and \(a = b = c = d\), we have:
    \[
    4a = 20 \implies a = 5
    \]

12. Therefore, \(a = b = c = d = 5\).

13. Verifying the second condition:
    \[
    ab + ac + ad + bc + bd + cd = 6 \cdot 5^2 = 6 \cdot 25 = 150
    \]

14. Both conditions are satisfied with \(a = b = c = d = 5\).

The final answer is \( \boxed{ (5, 5, 5, 5) } \)