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.
For the product to be positive, $x-\frac{2}{3}$ and $x+\frac{3}{2}$ have to be both negative or both positive. Consider the case when $x-\frac{2}{3}$ and $x+\frac{3}{2}$ are both negative.
$$x-\frac{2}{3}<0$$ $$x+\frac{3}{2}<0$$
The solution satisfying both inequalities is $x<-\frac{3}{2}$.
$$x<-\frac{3}{2}$$
Consider the case when $x-\frac{2}{3}$ and $x+\frac{3}{2}$ are both positive.
$$x+\frac{3}{2}>0$$ $$x-\frac{2}{3}>0$$
The solution satisfying both inequalities is $x>\frac{2}{3}$.
$$x>\frac{2}{3}$$
The final solution is the union of the obtained solutions.
