Let \(u = 2x - 3\). Then \(u^2 = (2x - 3)^2\).

Substitute into the equation:

\(u^2 - \frac{4}{u^2} - 3 = 0\).

Multiply through by \(u^2\) to eliminate the fraction:

\(u^4 - 3u^2 - 4 = 0\).

Let \(v = u^2\), then:

\(v^2 - 3v - 4 = 0\).

Factorize:

\((v - 4)(v + 1) = 0\).

So, \(v = 4\) or \(v = -1\).

Since \(v = u^2\), \(u^2 = 4\) or \(u^2 = -1\).

\(u^2 = -1\) has no real solutions.

For \(u^2 = 4\), \(u = \pm 2\).

Substitute back for \(u = 2x - 3\):

\(2x - 3 = 2\) or \(2x - 3 = -2\).

Solving these gives:

\(2x = 5\) or \(2x = 1\).

Thus, \(x = \frac{5}{2}\) or \(x = \frac{1}{2}\).