Solve m^2+12m<0

$$m ^ { 2 } + 12 m < 0$$

Solve for m

$$m\in \left(-12,0\right)$$

Solution Steps

Factor out m.

$$m\left(m+12\right)<0$$

For the product to be negative, m+12 and m have to be of the opposite signs. Consider the case when m+12 is positive and m is negative.

$$m+12>0 m<0$$

The solution satisfying both inequalities is m\in \left(-12,0\right).

$$m\in \left(-12,0\right)$$

Consider the case when m is positive and m+12 is negative.

$$m>0 m+12<0$$

This is false for any m.

$$m\in \emptyset$$

The final solution is the union of the obtained solutions.

$$m\in \left(-12,0\right)$$