Solve 15(x^2-1)>-16x

$$15 ( x ^ { 2 } - 1 ) > - 16 x$$

Solve for x

$$x\in \left(-\infty,-\frac{5}{3}\right)\cup \left(\frac{3}{5},\infty\right)$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 15 by x^{2}-1.

$$15x^{2}-15>-16x$$

Add 16x to both sides.

$$15x^{2}-15+16x>0$$

To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$15x^{2}-15+16x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 15 for a, 16 for b, and -15 for c in the quadratic formula.

$$x=\frac{-16±\sqrt{16^{2}-4\times 15\left(-15\right)}}{2\times 15}$$

Do the calculations.

$$x=\frac{-16±34}{30}$$

Solve the equation x=\frac{-16±34}{30} when ± is plus and when ± is minus.

$$x=\frac{3}{5} x=-\frac{5}{3}$$

Rewrite the inequality by using the obtained solutions.

$$15\left(x-\frac{3}{5}\right)\left(x+\frac{5}{3}\right)>0$$

For the product to be positive, x-\frac{3}{5} and x+\frac{5}{3} have to be both negative or both positive. Consider the case when x-\frac{3}{5} and x+\frac{5}{3} are both negative.

$$x-\frac{3}{5}<0 x+\frac{5}{3}<0$$

The solution satisfying both inequalities is x<-\frac{5}{3}.

$$x<-\frac{5}{3}$$

Consider the case when x-\frac{3}{5} and x+\frac{5}{3} are both positive.

$$x+\frac{5}{3}>0 x-\frac{3}{5}>0$$

The solution satisfying both inequalities is x>\frac{3}{5}.

$$x>\frac{3}{5}$$

The final solution is the union of the obtained solutions.

$$x<-\frac{5}{3}\text{; }x>\frac{3}{5}$$