Solve 3(x-2)(2x+1)>0

$$3 ( x - 2 ) ( 2 x + 1 ) > 0$$

Solve for x

$$x\in \left(-\infty,-\frac{1}{2}\right)\cup \left(2,\infty\right)$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3 by x-2.

$$\left(3x-6\right)\left(2x+1\right)>0$$

Use the distributive property to multiply 3x-6 by 2x+1 and combine like terms.

$$6x^{2}-9x-6>0$$

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.

$$6x^{2}-9x-6=0$$

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, -9 for b, and -6 for c in the quadratic formula.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 6\left(-6\right)}}{2\times 6}$$

Do the calculations.

$$x=\frac{9±15}{12}$$

Solve the equation x=\frac{9±15}{12} when ± is plus and when ± is minus.

$$x=2 x=-\frac{1}{2}$$

Rewrite the inequality by using the obtained solutions.

$$6\left(x-2\right)\left(x+\frac{1}{2}\right)>0$$

For the product to be positive, x-2 and x+\frac{1}{2} have to be both negative or both positive. Consider the case when x-2 and x+\frac{1}{2} are both negative.

$$x-2<0 x+\frac{1}{2}<0$$

The solution satisfying both inequalities is x<-\frac{1}{2}.

$$x<-\frac{1}{2}$$

Consider the case when x-2 and x+\frac{1}{2} are both positive.

$$x+\frac{1}{2}>0 x-2>0$$

The solution satisfying both inequalities is x>2.

$$x>2$$

The final solution is the union of the obtained solutions.

$$x<-\frac{1}{2}\text{; }x>2$$