Solve 12x^2-31x+20>0

$$12 x ^ { 2 } - 31 x + 20 > 0$$

Solve for x

$$x\in \left(-\infty,\frac{5}{4}\right)\cup \left(\frac{4}{3},\infty\right)$$

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.

$$12x^{2}-31x+20=0$$

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, -31 for b, and 20 for c in the quadratic formula.

$$x=\frac{-\left(-31\right)±\sqrt{\left(-31\right)^{2}-4\times 12\times 20}}{2\times 12}$$

Do the calculations.

$$x=\frac{31±1}{24}$$

Solve the equation x=\frac{31±1}{24} when ± is plus and when ± is minus.

$$x=\frac{4}{3} x=\frac{5}{4}$$

Rewrite the inequality by using the obtained solutions.

$$12\left(x-\frac{4}{3}\right)\left(x-\frac{5}{4}\right)>0$$

For the product to be positive, x-\frac{4}{3} and x-\frac{5}{4} have to be both negative or both positive. Consider the case when x-\frac{4}{3} and x-\frac{5}{4} are both negative.

$$x-\frac{4}{3}<0 x-\frac{5}{4}<0$$

The solution satisfying both inequalities is x<\frac{5}{4}.

$$x<\frac{5}{4}$$

Consider the case when x-\frac{4}{3} and x-\frac{5}{4} are both positive.

$$x-\frac{5}{4}>0 x-\frac{4}{3}>0$$

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

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

The final solution is the union of the obtained solutions.

$$x<\frac{5}{4}\text{; }x>\frac{4}{3}$$