Solve x^2+20x<0

$${ x }^{ 2 } +20x < 0$$

Solve for x

$$x\in \left(-20,0\right)$$

Solution Steps

Factor out x.

$$x\left(x+20\right)<0$$

For the product to be negative, x+20 and x have to be of the opposite signs. Consider the case when x+20 is positive and x is negative.

$$x+20>0 x<0$$

The solution satisfying both inequalities is x\in \left(-20,0\right).

$$x\in \left(-20,0\right)$$

Consider the case when x is positive and x+20 is negative.

$$x>0 x+20<0$$

This is false for any x.

$$x\in \emptyset$$

The final solution is the union of the obtained solutions.

$$x\in \left(-20,0\right)$$