Answer: \(x = \frac{3 + \sqrt{17}}{2} \text{ or } \frac{3 - \sqrt{17}}{2}\)
Compare with \(ax^2 + bx + c = 0\): \(a = 1\), \(b = -3\), \(c = -2\).
\(b^2 - 4ac = 9 + 8 = 17\)
\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
\[x = \frac{3 \pm \sqrt{17}}{2}\]