Solve (6-3d)(d-3)<0

$$( 6 - 3 d ) ( d - 3 ) < 0$$

Solve for d

$$d\in \left(-\infty,2\right)\cup \left(3,\infty\right)$$

Solution Steps

For the product to be negative, 6-3d and d-3 have to be of the opposite signs. Consider the case when 6-3d is positive and d-3 is negative.

$$6-3d>0 d-3<0$$

The solution satisfying both inequalities is d<2.

$$d<2$$

Consider the case when d-3 is positive and 6-3d is negative.

$$d-3>0 6-3d<0$$

The solution satisfying both inequalities is d>3.

$$d>3$$

The final solution is the union of the obtained solutions.

$$d<2\text{; }d>3$$