Given the Fibonacci sequence with $f_0=f_1=1$and for $n\geq 1, f_{n+1}=f_n+f_{n-1}$, find all real solutions to the equation: $$x^{2024}=f_{2023}x+f_{2022}.$$
1. **Base Case Verification:**
   We start by verifying the base case for \( n = 1 \) and \( n = 2 \).

   For \( n = 1 \):
   \[
   x^2 - f_1 x - f_0 = x^2 - x - 1
   \]
   This is indeed equal to \( (x^2 - x - 1)(f_0) \) since \( f_0 = 1 \).

   For \( n = 2 \):
   \[
   x^3 - f_2 x - f_1 = x^3 - 2x - 1
   \]
   We can factorize this as:
   \[
   x^3 - 2x - 1 = (x^2 - x - 1)(x + 1)
   \]
   This is true since \( f_2 = 2 \) and \( f_1 = 1 \).

2. **Inductive Step:**
   Assume the statement is true for \( n = k \):
   \[
   x^{k+1} - f_k x - f_{k-1} = (x^2 - x - 1)(f_0 x^{k-1} + f_1 x^{k-2} + \cdots + f_{k-1})
   \]
   We need to show it holds for \( n = k + 1 \):
   \[
   x^{k+2} - f_{k+1} x - f_k
   \]
   Using the inductive hypothesis:
   \[
   (x^2 - x - 1)(f_0 x^k + f_1 x^{k-1} + \cdots + f_k) = (x^2 - x - 1)(x(f_0 x^{k-1} + f_1 x^{k-2} + \cdots + f_{k-1}) + f_k)
   \]
   Simplifying the right-hand side:
   \[
   = x(x^{k+1} - f_k x - f_{k-1}) + f_k (x^2 - x - 1)
   \]
   \[
   = x^{k+2} - f_{k+1} x - f_k
   \]
   This completes the inductive step.

3. **Solving the Given Equation:**
   Given:
   \[
   x^{2024} = f_{2023} x + f_{2022}
   \]
   Using the proven formula:
   \[
   x^{2024} - f_{2023} x - f_{2022} = (x^2 - x - 1)(f_0 x^{2022} + f_1 x^{2021} + \cdots + f_{2022})
   \]
   For the equation to hold, the right-hand side must be zero:
   \[
   (x^2 - x - 1)(f_0 x^{2022} + f_1 x^{2021} + \cdots + f_{2022}) = 0
   \]
   Since \( f_0 x^{2022} + f_1 x^{2021} + \cdots + f_{2022} \neq 0 \) for all \( x \), we must have:
   \[
   x^2 - x - 1 = 0
   \]
   Solving this quadratic equation:
   \[
   x = \frac{1 \pm \sqrt{5}}{2}
   \]

The final answer is \( \boxed{\frac{1 \pm \sqrt{5}}{2}} \).