Find a real function $f : [0,\infty)\to \mathbb R$ such that $f(2x+1) = 3f(x)+5$, for all $x$ in $[0,\infty)$.
To find a real function \( f : [0,\infty) \to \mathbb{R} \) such that \( f(2x+1) = 3f(x) + 5 \) for all \( x \in [0,\infty) \), we can proceed as follows:

1. **Assume a form for \( f(x) \)**:
   Let's assume \( f(x) = c \) where \( c \) is a constant. Then, substituting into the functional equation:
   \[
   f(2x+1) = 3f(x) + 5
   \]
   becomes:
   \[
   c = 3c + 5
   \]

2. **Solve for \( c \)**:
   \[
   c = 3c + 5
   \]
   \[
   c - 3c = 5
   \]
   \[
   -2c = 5
   \]
   \[
   c = -\frac{5}{2}
   \]

3. **Verify the solution**:
   Substitute \( f(x) = -\frac{5}{2} \) back into the original functional equation to verify:
   \[
   f(2x+1) = -\frac{5}{2}
   \]
   \[
   3f(x) + 5 = 3\left(-\frac{5}{2}\right) + 5 = -\frac{15}{2} + 5 = -\frac{15}{2} + \frac{10}{2} = -\frac{5}{2}
   \]
   Both sides are equal, confirming that \( f(x) = -\frac{5}{2} \) is indeed a solution.

4. **Conclusion**:
   The function \( f(x) = -\frac{5}{2} \) satisfies the given functional equation for all \( x \in [0, \infty) \).

The final answer is \( \boxed{ f(x) = -\frac{5}{2} } \).