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

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

Solve for x

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

Steps Using the Quadratic Formula

Use the distributive property to multiply 3 by 2x-1.

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

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

$$3x+6x^{2}-3>0$$

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.

$$3x+6x^{2}-3=0$$

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, 3 for b, and -3 for c in the quadratic formula.

$$x=\frac{-3±\sqrt{3^{2}-4\times 6\left(-3\right)}}{2\times 6}$$

Do the calculations.

$$x=\frac{-3±9}{12}$$

Solve the equation x=\frac{-3±9}{12} when ± is plus and when ± is minus.

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

Rewrite the inequality by using the obtained solutions.

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

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

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

The solution satisfying both inequalities is x<-1.

$$x<-1$$

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

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

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

$$x>\frac{1}{2}$$

The final solution is the union of the obtained solutions.

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