Hence solve the equation \( 4 t^{2}+35 t-9=0 \)

To solve the equation \( 4t^{2} + 35t - 9 = 0 \), we can use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 4 \), \( b = 35 \), and \( c = -9 \). First, calculate the discriminant: \[ b^2 - 4ac = 35^2 - 4 \times 4 \times (-9) = 1225 + 144 = 1369. \] Now, take the square root of the discriminant: \[ \sqrt{1369} = 37. \] Now apply the quadratic formula: \[ t = \frac{-35 \pm 37}{2 \times 4}. \] This gives us two solutions: \[ t_1 = \frac{-35 + 37}{8} = \frac{2}{8} = \frac{1}{4}, \] \[ t_2 = \frac{-35 - 37}{8} = \frac{-72}{8} = -9. \] Thus, the solutions are \( t = \frac{1}{4} \) and \( t = -9 \).