Solve 2-3x-9x^2>0

$$2 - 3 x - 9 x ^ { 2 } > 0$$

Solve for x

$$x\in \left(-\frac{2}{3},\frac{1}{3}\right)$$

Steps Using the Quadratic Formula

Multiply the inequality by -1 to make the coefficient of the highest power in 2-3x-9x^{2} positive. Since -1 is negative, the inequality direction is changed.

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

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.

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

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 9 for a, 3 for b, and -2 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{-3±9}{18}$$

Solve the equation x=\frac{-3±9}{18} when ± is plus and when ± is minus.

$$x=\frac{1}{3} x=-\frac{2}{3}$$

Rewrite the inequality by using the obtained solutions.

$$9\left(x-\frac{1}{3}\right)\left(x+\frac{2}{3}\right)<0$$

For the product to be negative, x-\frac{1}{3} and x+\frac{2}{3} have to be of the opposite signs. Consider the case when x-\frac{1}{3} is positive and x+\frac{2}{3} is negative.

$$x-\frac{1}{3}>0 x+\frac{2}{3}<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x+\frac{2}{3} is positive and x-\frac{1}{3} is negative.

$$x+\frac{2}{3}>0 x-\frac{1}{3}<0$$

The solution satisfying both inequalities is x\in \left(-\frac{2}{3},\frac{1}{3}\right).

$$x\in \left(-\frac{2}{3},\frac{1}{3}\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(-\frac{2}{3},\frac{1}{3}\right)$$