1. Given the polynomial equation:
$$4x^4 - 3x^2 + 7x - 3 = 0$$
we need to find the sum of the real roots.

2. We start by using synthetic division to check for possible rational roots. Let's test $x = \frac{1}{2}$:
$$\text{Synthetic division of } 4x^4 - 3x^2 + 7x - 3 \text{ by } (2x - 1):$$
$$\begin{array}{r|rrrrr} 2x - 1 & 4 & 0 & -3 & 7 & -3 \\ \hline & 4 & 2 & -1 & 6 & 0 \\ \end{array}$$
The remainder is 0, so $x = \frac{1}{2}$ is a root, and $4x^4 - 3x^2 + 7x - 3$ can be factored as:
$$4x^4 - 3x^2 + 7x - 3 = (2x - 1)(2x^3 + x^2 - x + 3)$$

3. Next, we apply synthetic division again to the cubic polynomial $2x^3 + x^2 - x + 3$ to check for another rational root. Let's test $x = -\frac{3}{2}$:
$$\text{Synthetic division of } 2x^3 + x^2 - x + 3 \text{ by } (2x + 3):$$
$$\begin{array}{r|rrrr} 2x + 3 & 2 & 1 & -1 & 3 \\ \hline & 2 & -2 & 2 & 0 \\ \end{array}$$
The remainder is 0, so $x = -\frac{3}{2}$ is a root, and $2x^3 + x^2 - x + 3$ can be factored as:
$$2x^3 + x^2 - x + 3 = (2x + 3)(x^2 - x + 1)$$

4. Now, we have:
$$4x^4 - 3x^2 + 7x - 3 = (2x - 1)(2x + 3)(x^2 - x + 1)$$

5. The quadratic $x^2 - x + 1$ has no real roots because its discriminant is negative:
$$\Delta = (-1)^2 - 4 \cdot 1 \cdot 1 = 1 - 4 = -3$$

6. Therefore, the only real roots of the original polynomial are $x = \frac{1}{2}$ and $x = -\frac{3}{2}$.

7. The sum of the real roots is:
$$\frac{1}{2} + \left(-\frac{3}{2}\right) = \frac{1}{2} - \frac{3}{2} = -1$$

The final answer is $\boxed{-1}$.