The equation \( 12 t^{2}+13 t-4=0 \) has solutions of the form \[ t=\frac{N \pm \sqrt{D}}{M} \] (A) Use the quadratic formula to solve this equation and find the appropriate integer values of \( N, M \), and \( D \). Do not worry about simplifying the \( \sqrt{D} \) yet in this part of the problem. N \( N=\square \); \( D=\square \)

To solve the equation \( 12 t^{2}+13 t-4=0 \) using the quadratic formula, we first identify the coefficients: \( a = 12 \), \( b = 13 \), and \( c = -4 \). The quadratic formula is given by \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \] Let’s calculate: 1. \( -b = -13 \) 2. \( b^2 = 13^2 = 169 \) 3. \( 4ac = 4 \cdot 12 \cdot (-4) = -192 \) 4. Therefore, the discriminant \( D = b^2 - 4ac = 169 + 192 = 361 \). Now substituting back into the formula gives us: \[ t = \frac{-13 \pm \sqrt{361}}{2 \cdot 12} = \frac{-13 \pm 19}{24}. \] So, we have \( N = -13 \), \( M = 24 \), and \( D = 361 \).