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# What are the zeros of the quadratic function f(x) = 8x^{2} - 16x - 15?

**Solution:**

Given, f(x) = 8x^{2} - 16x - 15

We have to find zero of the quadratic function.

By using quadratic formula,

\(x = \frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\)

Here, a = 8, b = -16, c = -15

\(x=\frac{-(-16)\pm \sqrt{(-16)^{2}-4(8)(-15)}}{2(8)}\)

\(x=\frac{16\pm \sqrt{256+480}}{16}\)

\(x=\frac{16\pm \sqrt{736}}{16}\)

\(x=\frac{16\pm27.13}{16}\)

Now, \(x=\frac{16+27.13}{16}=\frac{43.13}{16}=2.69\)

\(x=\frac{16-27.13}{16}=\frac{-11.13}{16}=-0.69\)

Therefore, the zeros of the quadratic function are 2.69 and -0.69.

## What are the zeros of the quadratic function f(x) = 8x^{2} - 16x - 15?

**Summary:**

x = -0.69 and x = 2.69 are zeros of the quadratic function f(x) = 8x^{2} - 16x - 15.

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