-3x^2-7x+6>0
To solve the inequality \(-3x^2 - 7x + 6 > 0\), we will first rewrite it in a more standard form:
Multiply the entire inequality by \(-1\) to make the leading coefficient positive (remember to flip the inequality sign): \[ 3x^2 + 7x - 6 < 0 \]
Next, we will find the roots of the quadratic equation \(3x^2 + 7x - 6 = 0\) using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a = 3\), \(b = 7\), and \(c = -6\).
Calculate the discriminant: \[ b^2 - 4ac = 7^2 - 4(3)(-6) = 49 + 72 = 121 \]
Now, apply the quadratic formula: \[ x = \frac{-7 \pm \sqrt{121}}{2 \cdot 3} = \frac{-7 \pm 11}{6} \]
Find the two roots: \[ x_1 = \frac{-7 + 11}{6} = \frac{4}{6} = \frac{2}{3} \] \[ x_2 = \frac{-7 - 11}{6} = \frac{-18}{6} = -3 \]
The roots are \(x_1 = \frac{2}{3}\) and \(x_2 = -3\).
Now, we need to determine the intervals where the quadratic \(3x^2 + 7x - 6\) is less than zero. The roots divide the x-axis into three intervals:
- \((-\infty, -3)\)
- \((-3, \frac{2}{3})\)
- \((\frac{2}{3}, \infty)\)
We can test a point from each interval to see where the quadratic is negative:
- For the interval \((-\infty, -3)\), test \(x = -4\): \[ 3(-4)^2 + 7(-4) - 6 = 48 - 28 - 6 = 14 \quad (\text{positive}) \]
- For the interval \((-3, \frac{2}{3})\), test \(x = 0\): \[ 3(0)^2 + 7(0) - 6 = -6 \quad (\text{negative}) \]
- For the interval \((\frac{2}{3}, \infty)\), test \(x = 1\): \[ 3(1)^2 + 7(1) - 6 = 3 + 7 - 6 = 4 \quad (\text{positive}) \]
The quadratic is less than zero only in the interval \((-3, \frac{2}{3})\).
Therefore, the solution to the inequality \(-3x^2 - 7x + 6 > 0\) is: \[ \boxed{(-3, \frac{2}{3})} \]
