X2+3x-28<0

To solve the inequality \( x^2 + 3x - 28 < 0 \), we'll first find the roots of the equation \( x^2 + 3x - 28 = 0 \). We can use the quadratic formula, which is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] In this case, \( a = 1 \), \( b = 3 \), and \( c = -28 \). Plugging in these values: \[ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-28)}}{2 \cdot 1} \] Calculating the discriminant: \[ b^2 - 4ac = 9 + 112 = 121 \] Now plug the discriminant back into the formula: \[ x = \frac{-3 \pm \sqrt{121}}{2} = \frac{-3 \pm 11}{2} \] This gives us two roots: 1. \( x = \frac{-3 + 11}{2} = \frac{8}{2} = 4 \) 2. \( x = \frac{-3 - 11}{2} = \frac{-14}{2} = -7 \) Now we have the two roots \( x = 4 \) and \( x = -7 \). Next, we'll analyze the intervals defined by these roots: 1. \( (-\infty, -7) \) 2. \( (-7, 4) \) 3. \( (4, \infty) \) We'll pick test points from each interval to determine where the inequality \( x^2 + 3x - 28 < 0 \) holds. 1. **Test point in \( (-\infty, -7) \)**: Let’s choose \( x = -8 \): \[ (-8)^2 + 3(-8) - 28 = 64 - 24 - 28 = 12 \quad (\text{not } < 0) \] 2. **Test point in \( (-7, 4) \)**: Let’s choose \( x = 0 \): \[ 0^2 + 3(0) - 28 = -28 \quad (< 0) \] 3. **Test point in \( (4, \infty) \)**: Let’s choose \( x = 5 \): \[ 5^2 + 3(5) - 28 = 25 + 15 - 28 = 12 \quad (\text{not } < 0) \] From the test points, we find that the inequality \( x^2 + 3x - 28 < 0 \) holds in the interval \( (-7, 4) \). Thus, the solution to the inequality \( x^2 + 3x - 28 < 0 \) is: \[ \boxed{(-7, 4)} \]