To find the zeros of the quadratic equation \( x^2 - 16x - 13 = 4 \), we first move all terms to one side of the equation:
\[ x^2 - 16x - 13 - 4 = 0 \]
This simplifies to:
\[ x^2 - 16x - 17 = 0 \]
Next, we can use the quadratic formula to find the zeros of the equation, which states that for any quadratic equation of the form \( ax^2 + bx + c = 0 \), the solutions for \( x \) can be found using:
\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} \]
In our equation, \( a = 1 \), \( b = -16 \), and \( c = -17 \).
Calculating the discriminant:
\[ b^2 - 4ac = (-16)^2 - 4(1)(-17) = 256 + 68 = 324 \]
Now we apply the quadratic formula:
\[ x = \frac{{-(-16) \pm \sqrt{324}}}{{2 \cdot 1}} = \frac{{16 \pm 18}}{2} \]
Calculating the two possible values:
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When we take the positive root: \[ x = \frac{{16 + 18}}{2} = \frac{34}{2} = 17 \]
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When we take the negative root: \[ x = \frac{{16 - 18}}{2} = \frac{-2}{2} = -1 \]
Thus, the zeros of the quadratic equation \( x^2 - 16x - 17 = 0 \) are \( -1 \) and \( 17 \).
Since we need to enter the smaller value first, the final answer is:
\[ -1, 17 \]