The equation \( 12 t^{2}+13 t-4=0 \) has solutions of the form \[ t=\frac{N \pm \sqrt{D}}{M} \]

To solve the equation \( 12t^{2} + 13t - 4 = 0 \), we can use the quadratic formula, which is \( t = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \). Here, \( a = 12 \), \( b = 13 \), and \( c = -4 \). First, we calculate the discriminant \( D \): \[ D = b^{2} - 4ac = 13^{2} - 4 \cdot 12 \cdot (-4) = 169 + 192 = 361 \] Now we can plug the values back into the formula: \[ t = \frac{-13 \pm \sqrt{361}}{2 \cdot 12} = \frac{-13 \pm 19}{24} \] Therefore, the solutions are: \[ t = \frac{6}{24} \text{ and } t = \frac{-32}{24} \] Which simplifies to: \[ t = \frac{1}{4} \text{ and } t = -\frac{4}{3} \] So, we can express the solutions in the form \( t = \frac{N \pm \sqrt{D}}{M} \), where \( N = -13 \), \( D = 361 \), and \( M = 24 \).