To find the zeros of the quadratic equation \( x^2 - 16x - 13 = 4 \), we can first rearrange it into standard form:
\[ x^2 - 16x - 13 - 4 = 0 \]
This simplifies to:
\[ x^2 - 16x - 17 = 0 \]
Now we will use the quadratic formula, which is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = -16 \), and \( c = -17 \).
Now we calculate the discriminant:
\[ b^2 - 4ac = (-16)^2 - 4 \cdot 1 \cdot (-17) = 256 + 68 = 324 \]
Next, we find the square root of the discriminant:
\[ \sqrt{324} = 18 \]
Now we can apply the quadratic formula:
\[ x = \frac{-(-16) \pm 18}{2 \cdot 1} = \frac{16 \pm 18}{2} \]
Calculating the two possible values for \( x \):
- \( x_1 = \frac{16 + 18}{2} = \frac{34}{2} = 17 \)
- \( x_2 = \frac{16 - 18}{2} = \frac{-2}{2} = -1 \)
Thus, the solutions are \( x = -1 \) and \( x = 17 \).
To enter the smaller value first:
\[ x = -1, \quad x = 17 \]
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