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

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

Solve for x

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

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.

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

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

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

Do the calculations.

$$x=\frac{-8±10}{6}$$

Solve the equation x=\frac{-8±10}{6} when ± is plus and when ± is minus.

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

Rewrite the inequality by using the obtained solutions.

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

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

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

The solution satisfying both inequalities is x<-3.

$$x<-3$$

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

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

The solution satisfying both inequalities is x>\frac{1}{3}.

$$x>\frac{1}{3}$$

The final solution is the union of the obtained solutions.

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