Solve 6x^2-5x-6<0

$$6 x ^ { 2 } - 5 x - 6 < 0$$

Solve for x

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

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.

$$6x^{2}-5x-6=0$$

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 6 for a, -5 for b, and -6 for c in the quadratic formula.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 6\left(-6\right)}}{2\times 6}$$

Do the calculations.

$$x=\frac{5±13}{12}$$

Solve the equation x=\frac{5±13}{12} when ± is plus and when ± is minus.

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

Rewrite the inequality by using the obtained solutions.

$$6\left(x-\frac{3}{2}\right)\left(x+\frac{2}{3}\right)<0$$

For the product to be negative, x-\frac{3}{2} and x+\frac{2}{3} have to be of the opposite signs. Consider the case when x-\frac{3}{2} is positive and x+\frac{2}{3} is negative.

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

This is false for any x.

$$x\in \emptyset$$

Consider the case when x+\frac{2}{3} is positive and x-\frac{3}{2} is negative.

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

The solution satisfying both inequalities is x\in \left(-\frac{2}{3},\frac{3}{2}\right).

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

The final solution is the union of the obtained solutions.

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