Answer: \(x = \frac{2 + \sqrt{39}}{5} \text{ or } x = \frac{2 - \sqrt{39}}{5}\)
Half of \(-\frac{4}{5}\) is \(-\frac{2}{5}\); its square is \(\frac{4}{25}\).
\[5x^2 - 4x - 7 = 0\]\[x^2 - \frac{4}{5}x = \frac{7}{5} \quad (\div 5)\]\[x^2 - \frac{4}{5}x + \frac{4}{25} = \frac{7}{5} + \frac{4}{25}\]\[\left(x - \frac{2}{5}\right)^2 = \frac{39}{25}\]\[x - \frac{2}{5} = \pm\frac{\sqrt{39}}{5}\]\(x = \frac{2 + \sqrt{39}}{5}\) or \(x = \frac{2 - \sqrt{39}}{5}\).