Solve -5x^2+9x+2>0

$$- 5 x ^ { 2 } + 9 x + 2 > 0$$

Solve for x

$$x\in \left(-\frac{1}{5},2\right)$$

Steps Using the Quadratic Formula

Multiply the inequality by -1 to make the coefficient of the highest power in -5x^{2}+9x+2 positive. Since -1 is negative, the inequality direction is changed.

$$5x^{2}-9x-2<0$$

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.

$$5x^{2}-9x-2=0$$

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

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 5\left(-2\right)}}{2\times 5}$$

Do the calculations.

$$x=\frac{9±11}{10}$$

Solve the equation x=\frac{9±11}{10} when ± is plus and when ± is minus.

$$x=2 x=-\frac{1}{5}$$

Rewrite the inequality by using the obtained solutions.

$$5\left(x-2\right)\left(x+\frac{1}{5}\right)<0$$

For the product to be negative, x-2 and x+\frac{1}{5} have to be of the opposite signs. Consider the case when x-2 is positive and x+\frac{1}{5} is negative.

$$x-2>0 x+\frac{1}{5}<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x+\frac{1}{5} is positive and x-2 is negative.

$$x+\frac{1}{5}>0 x-2<0$$

The solution satisfying both inequalities is x\in \left(-\frac{1}{5},2\right).

$$x\in \left(-\frac{1}{5},2\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(-\frac{1}{5},2\right)$$