Answer: \(x = -1 \text{ or } x = -\frac{8}{5}\)
Compare with \(ax^2 + bx + c = 0\): \(a = 5\), \(b = 13\), \(c = 8\).
\(b^2 - 4ac = 169 - 160 = 9\)
\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
\[x = \frac{-13 \pm 3}{10}\]\[x = -1 \text{ or } x = -\frac{8}{5}\]