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:

\[ -\frac{7}{3} < x < 2 \]