Solve 3x^2+8x+4>0

$$3 x ^ { 2 } + 8 x + 4 > 0$$

Solve for x

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

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.

$$3x^{2}+8x+4=0$$

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 3 for a, 8 for b, and 4 for c in the quadratic formula.

$$x=\frac{-8±\sqrt{8^{2}-4\times 3\times 4}}{2\times 3}$$

Do the calculations.

$$x=\frac{-8±4}{6}$$

Solve the equation x=\frac{-8±4}{6} when ± is plus and when ± is minus.

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

Rewrite the inequality by using the obtained solutions.

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

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

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

The solution satisfying both inequalities is x<-2.

$$x<-2$$

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

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

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

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

The final solution is the union of the obtained solutions.

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