What are the zeros of the quadratic function f(x)=x^2-6x-16 ?

To find the zeros of the quadratic function \( f(x) = x^2 - 6x - 16 \), we need to solve the equation \( x^2 - 6x - 16 = 0 \).<br /><br />We can use the quadratic formula, which is given by:<br /><br />\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]<br /><br />For the quadratic equation \( ax^2 + bx + c = 0 \), the coefficients are:<br />- \( a = 1 \)<br />- \( b = -6 \)<br />- \( c = -16 \)<br /><br />Now, substitute these values into the quadratic formula:<br /><br />\[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-16)}}{2(1)} \]<br />\[ x = \frac{6 \pm \sqrt{36 + 64}}{2} \]<br />\[ x = \frac{6 \pm \sqrt{100}}{2} \]<br />\[ x = \frac{6 \pm 10}{2} \]<br /><br />This gives us two solutions:<br /><br />\[ x = \frac{6 + 10}{2} = \frac{16}{2} = 8 \]<br />\[ x = \frac{6 - 10}{2} = \frac{-4}{2} = -2 \]<br /><br />Therefore, the zeros of the quadratic function \( f(x) = x^2 - 6x - 16 \) are \( x = 8 \) and \( x = -2 \).