To find the zeros of the quadratic equation \( x^2 - 16x - 13 = 4 \), we first need to rearrange the equation into standard form:
- Subtract 4 from both sides:
\[ x^2 - 16x - 13 - 4 = 0 \]
which simplifies to:
\[ x^2 - 16x - 17 = 0. \]
Next, we can use the quadratic formula to find the values of \( x \):
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \]
where \( a = 1 \), \( b = -16 \), and \( c = -17 \).
Now we calculate \( b^2 - 4ac \):
\[ b^2 = (-16)^2 = 256, \] \[ 4ac = 4 \cdot 1 \cdot (-17) = -68. \]
Thus,
\[ b^2 - 4ac = 256 + 68 = 324. \]
Now substituting back into the quadratic formula:
\[ x = \frac{-(-16) \pm \sqrt{324}}{2 \cdot 1} = \frac{16 \pm 18}{2}. \]
This gives us two potential solutions:
\[ x = \frac{16 + 18}{2} = \frac{34}{2} = 17, \]
\[ x = \frac{16 - 18}{2} = \frac{-2}{2} = -1. \]
Therefore, the zeros of the equation are \( -1 \) and \( 17 \).
Since we are asked to enter the smaller value first, the final answer is:
\[ \boxed{-1}, 17. \]