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

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

Solve for x

$$x\in \left(-1,\frac{3}{2}\right)$$

Solution Steps

For the product to be negative, x+1 and 2x-3 have to be of the opposite signs. Consider the case when x+1 is positive and 2x-3 is negative.

$$x+1>0 2x-3<0$$

The solution satisfying both inequalities is x\in \left(-1,\frac{3}{2}\right).

$$x\in \left(-1,\frac{3}{2}\right)$$

Consider the case when 2x-3 is positive and x+1 is negative.

$$2x-3>0 x+1<0$$

This is false for any x.

$$x\in \emptyset$$

The final solution is the union of the obtained solutions.

$$x\in \left(-1,\frac{3}{2}\right)$$