\(4t^{2} + 7t - 1 = 0\) Using a = 4, b = 7, c = -1. \(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\) = \(\frac{-7 \pm \sqrt{7^{2} - 4(4)(-1)}}{2(4)}\) = \(\frac{-7 \pm \sqrt{49 + 16}}{8}\) = \(\frac{-7 \pm \sqrt{65}}{8}\) = \(\frac{-7 \pm 8.0623}{8}\) The positive answer = \(\frac{-7 + 8.0623}{8} = \frac{1.0623}{8}\) \(\approxeq 0.1328\) (4 decimal place)