Answer: \(x = \dfrac{3 + \sqrt{17}}{2}\) or \(x = \dfrac{3 - \sqrt{17}}{2}\)
\(a = 1\), \(b = -3\), \(c = -2\).
\(b^2 - 4ac = (-3)^2 - 4(1)(-2) = 9 + 8 = 17 > 0\).
\(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \dfrac{3 \pm \sqrt{17}}{2}\).
So \(x = \dfrac{3 + \sqrt{17}}{2}\) or \(x = \dfrac{3 - \sqrt{17}}{2}\).