Solve 12x^2-x-6>0

$$12 { x }^{ 2 } -x-6 > 0$$

Solve for x

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

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.

$$12x^{2}-x-6=0$$

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 12 for a, -1 for b, and -6 for c in the quadratic formula.

$$x=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 12\left(-6\right)}}{2\times 12}$$

Do the calculations.

$$x=\frac{1±17}{24}$$

Solve the equation x=\frac{1±17}{24} when ± is plus and when ± is minus.

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

Rewrite the inequality by using the obtained solutions.

$$12\left(x-\frac{3}{4}\right)\left(x+\frac{2}{3}\right)>0$$

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

$$x-\frac{3}{4}<0 x+\frac{2}{3}<0$$

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

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

Consider the case when x-\frac{3}{4} and x+\frac{2}{3} are both positive.

$$x+\frac{2}{3}>0 x-\frac{3}{4}>0$$

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

$$x>\frac{3}{4}$$

The final solution is the union of the obtained solutions.

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