Solve for all solutions of \( x \) \( x^{2}+6 x-15=0 \)
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Beyond the Answer
To solve the quadratic equation \( x^{2}+6x-15=0 \), we can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 1 \), \( b = 6 \), and \( c = -15 \). Plugging in these values: \[ x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot (-15)}}{2 \cdot 1} \] Calculating the discriminant: \[ 6^2 - 4 \cdot 1 \cdot (-15) = 36 + 60 = 96 \] Now substituting this into the formula: \[ x = \frac{-6 \pm \sqrt{96}}{2} \] Since \( \sqrt{96} = \sqrt{16 \cdot 6} = 4\sqrt{6} \), we have: \[ x = \frac{-6 \pm 4\sqrt{6}}{2} \] This simplifies to: \[ x = -3 \pm 2\sqrt{6} \] Therefore, the solutions for \( x \) are: \[ x = -3 + 2\sqrt{6} \quad \text{and} \quad x = -3 - 2\sqrt{6} \]
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