To find the zeros of the quadratic equation \( x^2 - 16x - 13 = 4 \), we start by rearranging the equation to standard form:
\[ x^2 - 16x - 13 - 4 = 0 \]
This simplifies to:
\[ x^2 - 16x - 17 = 0 \]
Next, we can use the quadratic formula, which is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In our equation \( x^2 - 16x - 17 = 0 \), we have:
- \( a = 1 \)
- \( b = -16 \)
- \( c = -17 \)
Now, we calculate the discriminant \( b^2 - 4ac \):
\[ b^2 - 4ac = (-16)^2 - 4(1)(-17) \] \[ = 256 + 68 \] \[ = 324 \]
Now we apply the quadratic formula:
\[ x = \frac{-(-16) \pm \sqrt{324}}{2 \cdot 1} \] \[ = \frac{16 \pm 18}{2} \]
This gives us two potential solutions:
- \( x = \frac{16 + 18}{2} = \frac{34}{2} = 17 \)
- \( x = \frac{16 - 18}{2} = \frac{-2}{2} = -1 \)
Thus, the zeros of the equation are:
\[ x = -1, \quad x = 17 \]
Finally, entering the smaller value first, we have:
\[ x = -1, \quad x = 17 \]