Comparing the equation \(3x^2 - 11x + 8 = 0\) with the standard form \(ax^2 + bx + c = 0\), we get: \(a = 3\), \(b = -11\), \(c = 8\) \[ \therefore b^2 - 4ac = (-11)^2 - 4 \times 3 \times 8 = 121 - 96 = 25 > 0 \] ∴ The equation \(3x^2 - 11x + 8 = 0\) has real roots. The roots are given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-(-11) \pm \sqrt{(-11)^2 - 4 \times 3 \times 8}}{2 \times 3} = \frac{11 \pm \sqrt{121 - 96}}{6} = \frac{11 \pm \sqrt{25}}{6} = \frac{11 \pm 5}{6} \] So, \[ x = \frac{11 + 5}{6} = \frac{16}{6} = \frac{8}{3} \quad \text{or} \quad x = \frac{11 - 5}{6} = \frac{6}{6} = 1 \] ∴ The required solutions are \(x = \frac{8}{3}\) and \(x = 1\).