Solve 2x+3x^2-11x-6<0

$$2 x + 3 x ^ { 2 } - 11 x - 6 < 0$$

Solve for x

$$x\in \left(\frac{3-\sqrt{17}}{2},\frac{\sqrt{17}+3}{2}\right)$$

Steps Using the Quadratic Formula

Combine 2x and -11x to get -9x.

$$-9x+3x^{2}-6<0$$

To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$-9x+3x^{2}-6=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 3 for a, -9 for b, and -6 for c in the quadratic formula.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 3\left(-6\right)}}{2\times 3}$$

Do the calculations.

$$x=\frac{9±3\sqrt{17}}{6}$$

Solve the equation x=\frac{9±3\sqrt{17}}{6} when ± is plus and when ± is minus.

$$x=\frac{\sqrt{17}+3}{2} x=\frac{3-\sqrt{17}}{2}$$

Rewrite the inequality by using the obtained solutions.

$$3\left(x-\frac{\sqrt{17}+3}{2}\right)\left(x-\frac{3-\sqrt{17}}{2}\right)<0$$

For the product to be negative, x-\frac{\sqrt{17}+3}{2} and x-\frac{3-\sqrt{17}}{2} have to be of the opposite signs. Consider the case when x-\frac{\sqrt{17}+3}{2} is positive and x-\frac{3-\sqrt{17}}{2} is negative.

$$x-\frac{\sqrt{17}+3}{2}>0 x-\frac{3-\sqrt{17}}{2}<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-\frac{3-\sqrt{17}}{2} is positive and x-\frac{\sqrt{17}+3}{2} is negative.

$$x-\frac{3-\sqrt{17}}{2}>0 x-\frac{\sqrt{17}+3}{2}<0$$

The solution satisfying both inequalities is x\in \left(\frac{3-\sqrt{17}}{2},\frac{\sqrt{17}+3}{2}\right).

$$x\in \left(\frac{3-\sqrt{17}}{2},\frac{\sqrt{17}+3}{2}\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(\frac{3-\sqrt{17}}{2},\frac{\sqrt{17}+3}{2}\right)$$