Solve. \[ r^{2}-6 r-7=0 \] \[ r= \] \( \square \) (Simplify your answer. Type an exact answer, using radicals as needed. Express complex numbers in terms of \( i \). Use a comma to separate answers as needed.)

To solve the quadratic equation \( r^{2} - 6r - 7 = 0 \), we can use the quadratic formula: \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 1 \), \( b = -6 \), and \( c = -7 \). Plugging in these values gives: \[ r = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-7)}}{2(1)} \] \[ r = \frac{6 \pm \sqrt{36 + 28}}{2} \] \[ r = \frac{6 \pm \sqrt{64}}{2} \] \[ r = \frac{6 \pm 8}{2} \] This results in two possible solutions: 1. \( r = \frac{14}{2} = 7 \) 2. \( r = \frac{-2}{2} = -1 \) Thus, the solutions for \( r \) are: \[ r = 7, -1 \]