Solve for \( x \) using the quadratic formula. \( \begin{array}{c}x^{2}-16 x+15=0 \\ x=[?], \\ \text { Enter the smallest solution first. } \\ \text { Quadratic Formula: } x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\end{array} \)
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To solve for \( x \) in the equation \( x^{2}-16 x+15=0 \) using the quadratic formula, we identify \( a = 1 \), \( b = -16 \), and \( c = 15 \). First, we'll calculate the discriminant \( b^{2}-4ac \): \[ b^{2} = (-16)^{2} = 256, \] \[ 4ac = 4 \cdot 1 \cdot 15 = 60, \] \[ b^{2}-4ac = 256-60 = 196. \] Now, we apply the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a} = \frac{-(-16) \pm \sqrt{196}}{2 \cdot 1} = \frac{16 \pm 14}{2}. \] This gives us two potential solutions: 1. \( x = \frac{16 + 14}{2} = \frac{30}{2} = 15 \). 2. \( x = \frac{16 - 14}{2} = \frac{2}{2} = 1 \). Thus, the solutions are \( x = 1 \) and \( x = 15 \). The smallest solution first is: \[ x = 1, \quad x = 15. \]