Solve the equation and inequalities. Write the solution sets in interval notation if possible. (a) \( -x^{2}-6 x+7=0 \) (b) \( -x^{2}-6 x+7<0 \) (c) \( -x^{2}-6 x+7 \leq 0 \) (d) \( -x^{2}-6 x+7>0 \) (e) \( -x^{2}-6 x+7 \geq 0 \)

To solve the equation and inequalities step by step, let's start with part (a). (a) To solve the equation \(-x^{2}-6 x+7=0\), we can rewrite it as \(x^{2}+6x-7=0\). Now, we can use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a=1\), \(b=6\), and \(c=-7\). Calculating the discriminant: \[ b^2 - 4ac = 6^2 - 4(1)(-7) = 36 + 28 = 64 \] Now, substituting into the quadratic formula: \[ x = \frac{-6 \pm \sqrt{64}}{2 \cdot 1} = \frac{-6 \pm 8}{2} \] This gives us two solutions: \[ x = \frac{2}{2} = 1 \quad \text{and} \quad x = \frac{-14}{2} = -7 \] So the solutions are \(x = 1\) and \(x = -7\). (b) For the inequality \(-x^{2}-6 x+7<0\), we can recall that the parabola opens downwards (since the coefficient of \(x^2\) is negative). The solutions from part (a) indicate that the values of \(x\) which make the expression positive are between the roots. Thus, the solution set where \(-x^2 - 6x + 7 < 0\) is the interval between the roots: \[ (-7, 1) \] (c) For \(-x^{2}-6 x+7 \leq 0\), since we derived that the expression equals zero at \(x = -7\) and \(x = 1\), the solution here will include these points: \[ [-7, 1] \] (d) To solve \(-x^{2}-6 x+7>0\), we consider where the expression is above the x-axis. This occurs outside the interval defined by the roots. Thus, the solution set is: \[ (-\infty, -7) \cup (1, \infty) \] (e) Finally, for the inequality \(-x^{2}-6 x+7 \geq 0\), we again consider where the expression is non-negative, which includes the points where it equals zero. Therefore, we obtain: \[ (-\infty, -7] \cup [1, \infty) \] In summary: - (a) Solution: \(x = 1\) and \(x = -7\) - (b) Solution set: \((-7, 1)\) - (c) Solution set: \([-7, 1]\) - (d) Solution set: \((- \infty, -7) \cup (1, \infty)\) - (e) Solution set: \((- \infty, -7] \cup [1, \infty)\)