Solve 2x^2+3x+1>0

$$2 { x }^{ 2 } +3x+1 > 0$$

Solve for x

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

Steps Using the Quadratic Formula

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.

$$2x^{2}+3x+1=0$$

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 1 for c in the quadratic formula.

$$x=\frac{-3±\sqrt{3^{2}-4\times 2\times 1}}{2\times 2}$$

Do the calculations.

$$x=\frac{-3±1}{4}$$

Solve the equation x=\frac{-3±1}{4} when ± is plus and when ± is minus.

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

Rewrite the inequality by using the obtained solutions.

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

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

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

The solution satisfying both inequalities is x<-1.

$$x<-1$$

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

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

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

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

The final solution is the union of the obtained solutions.

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