Find all $f: \mathbb R \to\mathbb R$ such that for all real numbers $x$, $f(x) \geq 0$ and for all real numbers $x$ and $y$, \[ f(x+y)+f(x-y)-2f(x)-2y^2=0.  \]
1. **Initial Condition and Simplification:**
   Given the functional equation:
   \[
   f(x+y) + f(x-y) - 2f(x) - 2y^2 = 0
   \]
   Let's start by plugging in \( y = 0 \):
   \[
   f(x+0) + f(x-0) - 2f(x) - 2 \cdot 0^2 = 0 \implies f(x) + f(x) - 2f(x) = 0 \implies 0 = 0
   \]
   This does not provide new information. Instead, let's consider the symmetry by plugging in \( y = x \):
   \[
   f(x+x) + f(x-x) - 2f(x) - 2x^2 = 0 \implies f(2x) + f(0) - 2f(x) - 2x^2 = 0
   \]
   Rearranging, we get:
   \[
   f(2x) = 2f(x) + 2x^2 - f(0)
   \]

2. **Breaking Down \( f(x) \) into Even and Odd Parts:**
   Let \( f(x) = h(x) + g(x) \) where \( h(x) \) is even and \( g(x) \) is odd. Thus:
   \[
   h(x) = h(-x) \quad \text{and} \quad g(x) = -g(-x)
   \]
   Substituting \( f(x) = h(x) + g(x) \) into the functional equation:
   \[
   h(x+y) + g(x+y) + h(x-y) + g(x-y) - 2h(x) - 2g(x) - 2y^2 = 0
   \]
   Since \( h \) is even and \( g \) is odd:
   \[
   h(x+y) + h(x-y) + g(x+y) + g(x-y) - 2h(x) - 2g(x) - 2y^2 = 0
   \]

3. **Analyzing the Even Part \( h(x) \):**
   By symmetry and the properties of even functions:
   \[
   h(x+y) + h(x-y) - 2h(x) = 2y^2
   \]
   Let \( h(x) = x^2 + c \) where \( c \) is a constant. Then:
   \[
   h(x+y) = (x+y)^2 + c = x^2 + 2xy + y^2 + c
   \]
   \[
   h(x-y) = (x-y)^2 + c = x^2 - 2xy + y^2 + c
   \]
   \[
   h(x+y) + h(x-y) = x^2 + 2xy + y^2 + c + x^2 - 2xy + y^2 + c = 2x^2 + 2y^2 + 2c
   \]
   \[
   2h(x) = 2(x^2 + c) = 2x^2 + 2c
   \]
   Thus:
   \[
   2x^2 + 2y^2 + 2c - 2x^2 - 2c = 2y^2
   \]
   This confirms \( h(x) = x^2 + c \).

4. **Analyzing the Odd Part \( g(x) \):**
   The remaining part of the functional equation:
   \[
   g(x+y) + g(x-y) = 2g(x)
   \]
   This is a Cauchy functional equation. For \( g(x) \) to be linear, we need to show it is continuous or bounded. Given \( f(x) \geq 0 \):
   \[
   h(x) + g(x) \geq 0 \implies x^2 + c + g(x) \geq 0
   \]
   \[
   g(x) \geq -x^2 - c
   \]
   Similarly, for \( g(-x) \):
   \[
   g(-x) = -g(x) \implies -g(x) \geq -x^2 - c \implies g(x) \leq x^2 + c
   \]
   Thus:
   \[
   -x^2 - c \leq g(x) \leq x^2 + c
   \]
   This implies \( g(x) \) is bounded and hence continuous. Therefore, \( g(x) = ax \) for some constant \( a \).

5. **Combining Results:**
   Combining \( h(x) \) and \( g(x) \):
   \[
   f(x) = x^2 + c + ax
   \]
   Given \( f(x) \geq 0 \):
   \[
   x^2 + ax + c \geq 0 \quad \forall x \in \mathbb{R}
   \]
   The quadratic \( x^2 + ax + c \) must have a non-positive discriminant:
   \[
   a^2 - 4c \leq 0
   \]

The final answer is \( \boxed{ f(x) = x^2 + ax + c } \) where \( a^2 - 4c \leq 0 \).