To find the zeros of the quadratic equation \( x^2 - 16x - 13 = 4 \), we first rearrange it into the standard form \( ax^2 + bx + c = 0 \).
- Subtract 4 from both sides:
\[ x^2 - 16x - 13 - 4 = 0 \] \[ x^2 - 16x - 17 = 0 \]
Now we can use the quadratic formula to solve for \( x \). The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In our equation, the coefficients are:
- \( a = 1 \)
- \( b = -16 \)
- \( c = -17 \)
- Next, we calculate the discriminant \( b^2 - 4ac \):
\[ b^2 - 4ac = (-16)^2 - 4(1)(-17) \] \[ = 256 + 68 \] \[ = 324 \]
- Now we'll use the quadratic formula to find the values of \( x \):
\[ x = \frac{-(-16) \pm \sqrt{324}}{2(1)} \] \[ x = \frac{16 \pm 18}{2} \]
- Now we calculate the two solutions:
- For the positive case:
\[ x = \frac{16 + 18}{2} = \frac{34}{2} = 17 \]
- For the negative case:
\[ x = \frac{16 - 18}{2} = \frac{-2}{2} = -1 \]
Thus, the solutions (or zeros) of the equation are \( -1 \) and \( 17 \).
The smaller of the two values is:
\[ \boxed{-1} \]
And the larger value is \( 17 \).