Solve 7^3x+left(13xright)^2-14x>0

$${ 7 }^{ 3 } x+ { \left(13x \right) }^{ 2 } -14x > 0$$

Solve for x

$$x\in \left(-\infty,-\frac{329}{169}\right)\cup \left(0,\infty\right)$$

Solution Steps

Calculate 7 to the power of 3 and get 343.

$$343x+\left(13x\right)^{2}-14x>0$$

Expand \left(13x\right)^{2}.

$$343x+13^{2}x^{2}-14x>0$$

Calculate 13 to the power of 2 and get 169.

$$343x+169x^{2}-14x>0$$

Combine 343x and -14x to get 329x.

$$329x+169x^{2}>0$$

Factor out x.

$$x\left(169x+329\right)>0$$

For the product to be positive, x+\frac{329}{169} and x have to be both negative or both positive. Consider the case when x+\frac{329}{169} and x are both negative.

$$x+\frac{329}{169}<0 x<0$$

The solution satisfying both inequalities is x<-\frac{329}{169}.

$$x<-\frac{329}{169}$$

Consider the case when x+\frac{329}{169} and x are both positive.

$$x>0 x+\frac{329}{169}>0$$

The solution satisfying both inequalities is x>0.

$$x>0$$

The final solution is the union of the obtained solutions.

$$x<-\frac{329}{169}\text{; }x>0$$