To solve the inequality \(|2x - 5| > 3|2x + 1|\), we consider two cases based on the definition of absolute value.

Case 1: \(2x - 5 > 3(2x + 1)\)

Simplifying, we get:

\(2x - 5 > 6x + 3\)

\(-5 - 3 > 6x - 2x\)

\(-8 > 4x\)

\(x < -2\)

Case 2: \(2x - 5 < -3(2x + 1)\)

Simplifying, we get:

\(2x - 5 < -6x - 3\)

\(2x + 6x < -3 + 5\)

\(8x < 2\)

\(x < \frac{1}{4}\)

Combining both cases, the solution is \(-2 < x < \frac{1}{4}\).