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

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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 $$D = 361$$.
$$ N = 13 $$, $$ M = 24 $$, and $$ D = 361 $$.

The equation has solutions of the form

(A) Use the quadratic formula to solve this equation and find the appropriate integer values of , and . Do not worry about simplifying the yet in this part of the problem.

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The equation has solutions of the form (A) Use the quadratic formula to solve this equation and find the appropriate integer values of , and . Do not worry about simplifying the yet in this part of the problem.