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. \[ \begin{array}{c} N=\begin{array}{c} 13 \times x ; D=361 \vee 0^{6} \\ M=24 \quad \checkmark \quad 0^{6} \end{array} \end{array} \]

Solve the equation \( 12t^2+13t-4=0 \). Solve the quadratic equation by following steps: - step0: Solve by factoring: \(12t^{2}+13t-4=0\) - step1: Factor the expression: \(\left(3t+4\right)\left(4t-1\right)=0\) - step2: Separate into possible cases: \(\begin{align}&4t-1=0\\&3t+4=0\end{align}\) - step3: Solve the equation: \(\begin{align}&t=\frac{1}{4}\\&t=-\frac{4}{3}\end{align}\) - step4: Rewrite: \(t_{1}=-\frac{4}{3},t_{2}=\frac{1}{4}\) The solutions to the equation \(12t^2 + 13t - 4 = 0\) are \(t_1 = -\frac{4}{3}\) and \(t_2 = \frac{1}{4}\). Now, we need to find the appropriate integer values of \(N\), \(M\), and \(D\) using the quadratic formula. The quadratic formula is given by: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] In this case, \(a = 12\), \(b = 13\), and \(c = -4\). Substituting these values into the quadratic formula, we get: \[ t = \frac{-13 \pm \sqrt{13^2 - 4(12)(-4)}}{2(12)} \] \[ t = \frac{-13 \pm \sqrt{169 + 192}}{24} \] \[ t = \frac{-13 \pm \sqrt{361}}{24} \] Therefore, \(N = 13\), \(M = 24\), and \(D = 361\).