Using the quadratic formula, we have \(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\) From the equation \(3x^{2} - 4x - 5 = 0\), a = 3, b = -4 and c = -5. \(\therefore x = \frac{-(-4) \pm \sqrt{(-4)^{2} - 4(3)(-5)}}{2(3)}\) = \(\frac{4 \pm \sqrt{16 + 60}}{6}\) = \(\frac{4 \pm \sqrt{76}}{6}\) = \(\frac{4 \pm 8.72}{6}\) = \(\frac{4 + 8.72}{6} \text{ or } \frac{4 - 8.72}{6}\) = \(\frac{12.72}{6} \text{ or } \frac{-4.72}{6}\) \(x = \text{2.12 or -0.79}\)