# How do you find the zeros of a function algebraically f(x)= x^2 - 14x - 4?

Jul 5, 2016

$x = 7 \pm \sqrt{53}$

#### Explanation:

The zeros of a function are defined as the point at which the value of the function is zero. We obtain these algebraically by setting the function equal to zero and solving the quadratic.

When we do this we get

${x}^{2} - 14 x - 4 = 0$

$x = \frac{14 \pm \sqrt{{\left(- 14\right)}^{2} - 4 \left(1\right) \left(- 4\right)}}{2} = \frac{14 \pm \sqrt{196 + 16}}{2}$
$x = \frac{14 \pm \sqrt{212}}{2} = \frac{14 \pm 2 \sqrt{53}}{2} = 7 \pm \sqrt{53}$