Solve -2x^2-5x-3<0

$$- 2 x ^ { 2 } - 5 x - 3 < 0$$

Solve for x

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

Steps Using the Quadratic Formula

Multiply the inequality by -1 to make the coefficient of the highest power in -2x^{2}-5x-3 positive. Since -1 is negative, the inequality direction is changed.

$$2x^{2}+5x+3>0$$

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.

$$2x^{2}+5x+3=0$$

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, 5 for b, and 3 for c in the quadratic formula.

$$x=\frac{-5±\sqrt{5^{2}-4\times 2\times 3}}{2\times 2}$$

Do the calculations.

$$x=\frac{-5±1}{4}$$

Solve the equation x=\frac{-5±1}{4} when ± is plus and when ± is minus.

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

Rewrite the inequality by using the obtained solutions.

$$2\left(x+1\right)\left(x+\frac{3}{2}\right)>0$$

For the product to be positive, x+1 and x+\frac{3}{2} have to be both negative or both positive. Consider the case when x+1 and x+\frac{3}{2} are both negative.

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

The solution satisfying both inequalities is x<-\frac{3}{2}.

$$x<-\frac{3}{2}$$

Consider the case when x+1 and x+\frac{3}{2} are both positive.

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

The solution satisfying both inequalities is x>-1.

$$x>-1$$

The final solution is the union of the obtained solutions.

$$x<-\frac{3}{2}\text{; }x>-1$$