Find a function $f$ such that $f(t^2 +t +1) = t$ for all real $t \ge 0$
1. We start with the given functional equation:
   \[
   f(t^2 + t + 1) = t \quad \text{for all real } t \ge 0.
   \]

2. Let \( m = t^2 + t + 1 \). We need to express \( t \) in terms of \( m \).

3. Solve the quadratic equation \( t^2 + t + 1 = m \) for \( t \):
   \[
   t^2 + t + 1 = m \implies t^2 + t + (1 - m) = 0.
   \]

4. Use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = 1 \), and \( c = 1 - m \):
   \[
   t = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (1 - m)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 - 4 + 4m}}{2} = \frac{-1 \pm \sqrt{4m - 3}}{2}.
   \]

5. Since \( t \ge 0 \), we take the positive root:
   \[
   t = \frac{-1 + \sqrt{4m - 3}}{2}.
   \]

6. Therefore, the function \( f \) that satisfies the given condition is:
   \[
   f(m) = \frac{-1 + \sqrt{4m - 3}}{2}.
   \]

The final answer is \( \boxed{ f(m) = \frac{-1 + \sqrt{4m - 3}}{2} } \).