To find the zeros of the quadratic equation \( x^2 - 16x - 13 = 4 \), we first rewrite the equation in standard form by moving all terms to one side:
\[ x^2 - 16x - 13 - 4 = 0 \] \[ x^2 - 16x - 17 = 0 \]
Next, we can use the quadratic formula to find the values of \( x \). The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In our equation \( a = 1 \), \( b = -16 \), and \( c = -17 \). We can now calculate the discriminant (\( b^2 - 4ac \)):
\[ b^2 - 4ac = (-16)^2 - 4 \cdot 1 \cdot (-17) \] \[ = 256 + 68 \] \[ = 324 \]
Now we substitute the values into the quadratic formula:
\[ x = \frac{-(-16) \pm \sqrt{324}}{2 \cdot 1} \] \[ = \frac{16 \pm 18}{2} \]
This gives us two solutions:
- \( x = \frac{16 + 18}{2} = \frac{34}{2} = 17 \)
- \( x = \frac{16 - 18}{2} = \frac{-2}{2} = -1 \)
Thus, the zeros of the quadratic equation are \( -1 \) and \( 17 \).
Since we need to enter the smaller value first, the answer is:
\[ -1, 17 \]
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