What are the zeros of the quadratic function $$ f(x)=2 x^{2}-10 x-3 $$ ? $$ x=-\frac{5}{2}-\frac{\sqrt{31}}{2} $$ and $$ x=-\frac{5}{2}+\frac{\sqrt{31}}{2} $$ $$ x=-\frac{5}{2}-\sqrt{\frac{37}{8}} $$ and $$ x=-\frac{5}{2}+\sqrt{\frac{37}{8}} $$ $$ x=\frac{5}{2}-\frac{\sqrt{31}}{2} $$ and $$ x=\frac{5}{2}+\frac{\sqrt{31}}{2} $$ $$ x=\frac{5}{2}-\sqrt{\frac{37}{8}} $$ and $$ x=\frac{5}{2}+\sqrt{\frac{37}{8}} $$

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We start with the quadratic function
$$
f(x)=2 x^{2}-10 x-3.
$$
To find the zeros, we set $$ f(x)=0 $$:
$$
2x^2 - 10x - 3 = 0.
$$

The quadratic formula states that, for an equation
$$
ax^2 + bx + c = 0,
$$
the solutions are
$$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.
$$

Here, $$ a = 2 $$, $$ b = -10 $$, and $$ c = -3 $$. Substitute these values into the quadratic formula:
$$
x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(2)(-3)}}{2 \cdot 2}.
$$

Simplify the expression:
- Compute $$-(-10)$$:
  $$
  -(-10) = 10.
  $$
- Compute the discriminant:
  $$
  (-10)^2 - 4(2)(-3) = 100 + 24 = 124.
  $$
- So, the formula becomes:
  $$
  x = \frac{10 \pm \sqrt{124}}{4}.
  $$

Now, simplify the square root. Notice that:
$$
\sqrt{124} = \sqrt{4 \cdot 31} = 2\sqrt{31}.
$$
Thus, the expression for $$ x $$ is:
$$
x = \frac{10 \pm 2\sqrt{31}}{4} = \frac{10}{4} \pm \frac{2\sqrt{31}}{4} = \frac{5}{2} \pm \frac{\sqrt{31}}{2}.
$$

Therefore, the zeros of the quadratic function are:
$$
x = \frac{5}{2} - \frac{\sqrt{31}}{2} \quad \text{and} \quad x = \frac{5}{2} + \frac{\sqrt{31}}{2}.
$$

Comparing this result with the given options, the correct choice is:
$$
x=\frac{5}{2}-\frac{\sqrt{31}}{2} \quad \text{and} \quad x=\frac{5}{2}+\frac{\sqrt{31}}{2}.
$$
The zeros of the quadratic function $$ f(x)=2x^{2}-10x-3 $$ are $$ x=\frac{5}{2}-\frac{\sqrt{31}}{2} $$ and $$ x=\frac{5}{2}+\frac{\sqrt{31}}{2} $$.

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