Solve 3x^2+12x-4>0

$$3 x ^ { 2 } + 12 x - 4 > 0$$

Solve for x

$$x\in \left(-\infty,-\frac{4\sqrt{3}}{3}-2\right)\cup \left(\frac{4\sqrt{3}}{3}-2,\infty\right)$$

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.

$$3x^{2}+12x-4=0$$

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, 12 for b, and -4 for c in the quadratic formula.

$$x=\frac{-12±\sqrt{12^{2}-4\times 3\left(-4\right)}}{2\times 3}$$

Do the calculations.

$$x=\frac{-12±8\sqrt{3}}{6}$$

Solve the equation x=\frac{-12±8\sqrt{3}}{6} when ± is plus and when ± is minus.

$$x=\frac{4\sqrt{3}}{3}-2 x=-\frac{4\sqrt{3}}{3}-2$$

Rewrite the inequality by using the obtained solutions.

$$3\left(x-\left(\frac{4\sqrt{3}}{3}-2\right)\right)\left(x-\left(-\frac{4\sqrt{3}}{3}-2\right)\right)>0$$

For the product to be positive, x-\left(\frac{4\sqrt{3}}{3}-2\right) and x-\left(-\frac{4\sqrt{3}}{3}-2\right) have to be both negative or both positive. Consider the case when x-\left(\frac{4\sqrt{3}}{3}-2\right) and x-\left(-\frac{4\sqrt{3}}{3}-2\right) are both negative.

$$x-\left(\frac{4\sqrt{3}}{3}-2\right)<0 x-\left(-\frac{4\sqrt{3}}{3}-2\right)<0$$

The solution satisfying both inequalities is x<-\frac{4\sqrt{3}}{3}-2.

$$x<-\frac{4\sqrt{3}}{3}-2$$

Consider the case when x-\left(\frac{4\sqrt{3}}{3}-2\right) and x-\left(-\frac{4\sqrt{3}}{3}-2\right) are both positive.

$$x-\left(-\frac{4\sqrt{3}}{3}-2\right)>0 x-\left(\frac{4\sqrt{3}}{3}-2\right)>0$$

The solution satisfying both inequalities is x>\frac{4\sqrt{3}}{3}-2.

$$x>\frac{4\sqrt{3}}{3}-2$$

The final solution is the union of the obtained solutions.

$$x<-\frac{4\sqrt{3}}{3}-2\text{; }x>\frac{4\sqrt{3}}{3}-2$$