9) \( r^{2}-4 r-91=7 \)

To solve the quadratic equation: \[ r^{2} - 4r - 91 = 7 \] **Step 1: Move all terms to one side to set the equation to zero.** \[ r^{2} - 4r - 91 - 7 = 0 \\ r^{2} - 4r - 98 = 0 \] **Step 2: Use the quadratic formula to solve for \( r \).** The quadratic formula is: \[ r = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \] For the equation \( r^{2} - 4r - 98 = 0 \): - \( a = 1 \) - \( b = -4 \) - \( c = -98 \) **Step 3: Plug the values into the quadratic formula.** \[ r = \frac{-(-4) \pm \sqrt{(-4)^{2} - 4(1)(-98)}}{2(1)} \\ r = \frac{4 \pm \sqrt{16 + 392}}{2} \\ r = \frac{4 \pm \sqrt{408}}{2} \] **Step 4: Simplify the square root and the expression.** \[ \sqrt{408} = \sqrt{4 \times 102} = 2\sqrt{102} \] So, \[ r = \frac{4 \pm 2\sqrt{102}}{2} \\ r = 2 \pm \sqrt{102} \] **Final Answer:** \[ r = 2 + \sqrt{102} \quad \text{or} \quad r = 2 - \sqrt{102} \] **Approximate Values:** If you need decimal approximations: \[ \sqrt{102} \approx 10.10 \] So, \[ r \approx 2 + 10.10 = 12.10 \\ \text{or} \\ r \approx 2 - 10.10 = -8.10 \] **Summary:** The solutions to the equation \( r^{2} - 4r - 91 = 7 \) are: \[ r = 2 + \sqrt{102} \quad \text{and} \quad r = 2 - \sqrt{102} \]