Solve the equation \[ x^{2}=5 x+14 \]
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To solve the equation \( x^{2} = 5x + 14 \), we first rearrange it to standard quadratic form: \[ x^{2} - 5x - 14 = 0 \] Next, we can either factor the equation, use the quadratic formula, or complete the square. Here, let's apply the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] In our equation, \( a = 1 \), \( b = -5 \), and \( c = -14 \). Plugging these values into the formula gives: \[ x = \frac{-(-5) \pm \sqrt{(-5)^{2} - 4 \cdot 1 \cdot (-14)}}{2 \cdot 1} \] Calculating the discriminant: \[ b^2 - 4ac = 25 + 56 = 81 \] So we have: \[ x = \frac{5 \pm \sqrt{81}}{2} \] Since \( \sqrt{81} = 9 \), our equation becomes: \[ x = \frac{5 \pm 9}{2} \] This leads to two potential solutions: 1. \( x = \frac{5 + 9}{2} = \frac{14}{2} = 7 \) 2. \( x = \frac{5 - 9}{2} = \frac{-4}{2} = -2 \) Therefore, the solutions to the equation are: \[ x = 7 \quad \text{and} \quad x = -2 \]