What is the sum of the real roots of the equation $4x^4-3x^2+7x-3=0$?

$ 
\textbf{(A)}\ -1
\qquad\textbf{(B)}\ -2
\qquad\textbf{(C)}\ -3
\qquad\textbf{(D)}\ -4
\qquad\textbf{(E)}\ \text {None of above}
$
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 \(\boxed{-1}\).