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

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

Solve for x

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

Steps Using the Quadratic Formula

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

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

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.

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

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

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

Do the calculations.

$$x=\frac{3±5}{4}$$

Solve the equation x=\frac{3±5}{4} when ± is plus and when ± is minus.

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

Rewrite the inequality by using the obtained solutions.

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

For the product to be positive, x-2 and x+\frac{1}{2} have to be both negative or both positive. Consider the case when x-2 and x+\frac{1}{2} are both negative.

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

The solution satisfying both inequalities is x<-\frac{1}{2}.

$$x<-\frac{1}{2}$$

Consider the case when x-2 and x+\frac{1}{2} are both positive.

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

The solution satisfying both inequalities is x>2.

$$x>2$$

The final solution is the union of the obtained solutions.

$$x<-\frac{1}{2}\text{; }x>2$$