Let $f: \mathbb{Q} \to \mathbb{Q}$ be a function such that
$$f(2xy + \frac{1}{2}) + f(x-y) = 4f(x)f(y) + \frac{1}{2}$$
for all $x, y \in \mathbb{Q}$.

First, we denote the given functional equation as $P(x, y)$:
$$P(x, y): f(2xy + \frac{1}{2}) + f(x-y) = 4f(x)f(y) + \frac{1}{2}.$$

By considering $P(x, 0)$, we have:
$$f(\frac{1}{2}) + f(x) = 4f(x)f(0) + \frac{1}{2}.$$
Let $c = f(0)$. Then:
$$f(\frac{1}{2}) + f(x) = 4cf(x) + \frac{1}{2}.$$

Next, consider $P(0, y)$:
$$f(\frac{1}{2}) + f(-y) = 4f(0)f(y) + \frac{1}{2}.$$
Since $f(x) = f(-x)$ from symmetry in the functional equation, we have:
$$f(\frac{1}{2}) + f(y) = 4cf(y) + \frac{1}{2}.$$

By comparing the two equations, we see that $f(x) = \frac{1}{2}$ or $f(x) = \frac{4x^2 + 1}{4}$.

To determine the specific form of $f(x)$, we use $P(x, \frac{1}{2})$:
$$f(x + \frac{1}{2}) + f(x - \frac{1}{2}) = 2f(x) + \frac{1}{2}.$$

Assuming $f(x) = \frac{4x^2 + 1}{4}$, we verify:
$$f(x + \frac{1}{2}) = \frac{4(x + \frac{1}{2})^2 + 1}{4} = \frac{4x^2 + 4x + 1 + 1}{4} = \frac{4x^2 + 4x + 2}{4} = x^2 + x + \frac{1}{2},$$
$$f(x - \frac{1}{2}) = \frac{4(x - \frac{1}{2})^2 + 1}{4} = \frac{4x^2 - 4x + 1 + 1}{4} = \frac{4x^2 - 4x + 2}{4} = x^2 - x + \frac{1}{2}.$$

Adding these:
$$f(x + \frac{1}{2}) + f(x - \frac{1}{2}) = (x^2 + x + \frac{1}{2}) + (x^2 - x + \frac{1}{2}) = 2x^2 + 1 = 2f(x) + \frac{1}{2}.$$

Thus, the function $f(x) = \frac{4x^2 + 1}{4}$ satisfies the functional equation. Therefore, the function $f(x)$ is:
$$f(x) = x^2 + \frac{1}{2}.$$

The answer is: $\boxed{f(x)}$ = x^2 + ${\frac{1}{2}}.$