To solve the quadratic equation \( 3x^2 - x\sqrt{3} + \sqrt{3} - 3 = 0 \), we can apply the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, the coefficients are:
- \( a = 3 \)
- \( b = -\sqrt{3} \)
- \( c = \sqrt{3} - 3 \)

Now calculate \( b^2 - 4ac \):

1. **Calculate \( b^2 \)**:
\[
b^2 = (-\sqrt{3})^2 = 3
\]

2. **Calculate \( 4ac \)**:
\[
4ac = 4 \cdot 3 \cdot (\sqrt{3} - 3) = 12(\sqrt{3} - 3) = 12\sqrt{3} - 36
\]

3. **Now calculate the discriminant \( b^2 - 4ac \)**:
\[
b^2 - 4ac = 3 - (12\sqrt{3} - 36) = 3 + 36 - 12\sqrt{3} = 39 - 12\sqrt{3}
\]

Now substitute these results back into the quadratic formula:

\[
x = \frac{-(-\sqrt{3}) \pm \sqrt{39 - 12\sqrt{3}}}{2 \cdot 3} = \frac{\sqrt{3} \pm \sqrt{39 - 12\sqrt{3}}}{6}
\]

Thus, the solutions for \( x \) are given by:

\[
x = \frac{\sqrt{3} + \sqrt{39 - 12\sqrt{3}}}{6} \quad \text{and} \quad x = \frac{\sqrt{3} - \sqrt{39 - 12\sqrt{3}}}{6}
\]

These represent the two possible values of \( x \).