3х2+х-14<0

To solve the inequality \( 3x^2 + x - 14 < 0 \), we first need to find the roots of the corresponding quadratic equation \( 3x^2 + x - 14 = 0 \). We can use the quadratic formula, which is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 3 \), \( b = 1 \), and \( c = -14 \). Let's compute the discriminant: \[ b^2 - 4ac = 1^2 - 4 \cdot 3 \cdot (-14) = 1 + 168 = 169 \] Now, applying the quadratic formula: \[ x = \frac{-1 \pm \sqrt{169}}{2 \cdot 3} = \frac{-1 \pm 13}{6} \] This gives us two solutions: \[ x_1 = \frac{12}{6} = 2 \] \[ x_2 = \frac{-14}{6} = -\frac{7}{3} \] Now we have the roots \( x_1 = 2 \) and \( x_2 = -\frac{7}{3} \). Next, we need to determine the intervals where the quadratic \( 3x^2 + x - 14 \) is less than 0. The roots divide the number line into three intervals: 1. \( (-\infty, -\frac{7}{3}) \) 2. \( (-\frac{7}{3}, 2) \) 3. \( (2, \infty) \) We can test a point from each interval to see where the expression is negative: 1. **Interval \( (-\infty, -\frac{7}{3}) \)**: Test \( x = -3 \) \[ 3(-3)^2 + (-3) - 14 = 27 - 3 - 14 = 10 \quad (>0) \] 2. **Interval \( (-\frac{7}{3}, 2) \)**: Test \( x = 0 \) \[ 3(0)^2 + 0 - 14 = -14 \quad (<0) \] 3. **Interval \( (2, \infty) \)**: Test \( x = 3 \) \[ 3(3)^2 + 3 - 14 = 27 + 3 - 14 = 16 \quad (>0) \] From these tests, we find that the quadratic is negative only in the interval \( (-\frac{7}{3}, 2) \). Thus, the solution to the inequality \( 3x^2 + x - 14 < 0 \) is: \[ -\frac{7}{3} < x < 2 \]