For the following equation, state the value of the discriminant and the number of real solutions. $$ \begin{array}{l}16 x^{2}-3 x-14=0\end{array} $$

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Solve the equation $$ 16x^2-3x-14=0 $$.
Solve the quadratic equation by following steps:
- step0: Solve using the quadratic formula:
  $$16x^{2}-3x-14=0$$
- step1: Solve using the quadratic formula:
  $$x=\frac{3\pm \sqrt{\left(-3\right)^{2}-4\times 16\left(-14\right)}}{2\times 16}$$
- step2: Simplify the expression:
  $$x=\frac{3\pm \sqrt{\left(-3\right)^{2}-4\times 16\left(-14\right)}}{32}$$
- step3: Simplify the expression:
  $$x=\frac{3\pm \sqrt{905}}{32}$$
- step4: Separate into possible cases:
  $$\begin{align}&x=\frac{3+\sqrt{905}}{32}\&x=\frac{3-\sqrt{905}}{32}\end{align}$$
- step5: Rewrite:
  $$x_{1}=\frac{3-\sqrt{905}}{32},x_{2}=\frac{3+\sqrt{905}}{32}$$
The solutions to the equation $$16x^2 - 3x - 14 = 0$$ are $$x_1 = \frac{3 - \sqrt{905}}{32}$$ and $$x_2 = \frac{3 + \sqrt{905}}{32}$$.

To find the discriminant, we use the formula $$D = b^2 - 4ac$$, where $$a = 16$$, $$b = -3$$, and $$c = -14$$.

Substitute the values into the formula:
$$D = (-3)^2 - 4(16)(-14)$$
$$D = 9 + 896$$
$$D = 905$$

Therefore, the discriminant is 905.

Since the discriminant is positive, the equation has two real solutions.
The discriminant is 905, and there are two real solutions.

For the following equation, state the value of the discriminant and the number of real solutions. \[ \begin{array}{l}16 x^{2}-3 x-14=0\end{array} \]

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