Solve y^2-4y^2-4y<0

$${ y }^{ 2 } -4 { y }^{ 2 } -4y < 0$$

Solve for y

$$y\in \left(-\infty,-\frac{4}{3}\right)\cup \left(0,\infty\right)$$

Solution Steps

Combine y^{2} and -4y^{2} to get -3y^{2}.

$$-3y^{2}-4y<0$$

Multiply the inequality by -1 to make the coefficient of the highest power in -3y^{2}-4y positive. Since -1 is negative, the inequality direction is changed.

$$3y^{2}+4y>0$$

Factor out y.

$$y\left(3y+4\right)>0$$

For the product to be positive, y+\frac{4}{3} and y have to be both negative or both positive. Consider the case when y+\frac{4}{3} and y are both negative.

$$y+\frac{4}{3}<0 y<0$$

The solution satisfying both inequalities is y<-\frac{4}{3}.

$$y<-\frac{4}{3}$$

Consider the case when y+\frac{4}{3} and y are both positive.

$$y>0 y+\frac{4}{3}>0$$

The solution satisfying both inequalities is y>0.

$$y>0$$

The final solution is the union of the obtained solutions.

$$y<-\frac{4}{3}\text{; }y>0$$