Use interval notation to write the solution sel of the inequality \[ x^{2}+5 x-14 \leq 0 \] The solution set is \( \square \). (Type your answer in interval notation.)

To solve the inequality \( x^2 + 5x - 14 \leq 0 \), we first find the roots of the corresponding equation \( x^2 + 5x - 14 = 0 \) using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-5 \pm \sqrt{5^2 - 4(1)(-14)}}{2(1)} = \frac{-5 \pm \sqrt{25 + 56}}{2} = \frac{-5 \pm \sqrt{81}}{2} = \frac{-5 \pm 9}{2}. \] This gives us the roots \( x = 2 \) and \( x = -7 \). Next, we analyze the sign of the quadratic \( x^2 + 5x - 14 \) on the intervals determined by these roots: \( (-\infty, -7) \), \( (-7, 2) \), and \( (2, \infty) \). Testing a point in each interval: 1. For \( x = -8 \) (in \( (-\infty, -7) \)): \( (-8)^2 + 5(-8) - 14 = 64 - 40 - 14 = 10 > 0 \). 2. For \( x = 0 \) (in \( (-7, 2) \)): \( 0^2 + 5(0) - 14 = -14 < 0 \). 3. For \( x = 3 \) (in \( (2, \infty) \)): \( 3^2 + 5(3) - 14 = 9 + 15 - 14 = 10 > 0 \). From this analysis, the quadratic is negative (or zero) on the interval \( [-7, 2] \). Thus, the solution set is \([-7, 2]\).