Solve (9x^2-16)-(3x-4)(2-5x)<0

$$( 9 x ^ { 2 } - 16 ) - ( 3 x - 4 ) ( 2 - 5 x ) < 0$$

Solve for x

$$x\in \left(-\frac{1}{4},\frac{4}{3}\right)$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3x-4 by 2-5x and combine like terms.

$$9x^{2}-16-\left(26x-15x^{2}-8\right)<0$$

To find the opposite of 26x-15x^{2}-8, find the opposite of each term.

$$9x^{2}-16-26x+15x^{2}+8<0$$

Combine 9x^{2} and 15x^{2} to get 24x^{2}.

$$24x^{2}-16-26x+8<0$$

Add -16 and 8 to get -8.

$$24x^{2}-8-26x<0$$

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.

$$24x^{2}-8-26x=0$$

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 24 for a, -26 for b, and -8 for c in the quadratic formula.

$$x=\frac{-\left(-26\right)±\sqrt{\left(-26\right)^{2}-4\times 24\left(-8\right)}}{2\times 24}$$

Do the calculations.

$$x=\frac{26±38}{48}$$

Solve the equation x=\frac{26±38}{48} when ± is plus and when ± is minus.

$$x=\frac{4}{3} x=-\frac{1}{4}$$

Rewrite the inequality by using the obtained solutions.

$$24\left(x-\frac{4}{3}\right)\left(x+\frac{1}{4}\right)<0$$

For the product to be negative, x-\frac{4}{3} and x+\frac{1}{4} have to be of the opposite signs. Consider the case when x-\frac{4}{3} is positive and x+\frac{1}{4} is negative.

$$x-\frac{4}{3}>0 x+\frac{1}{4}<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x+\frac{1}{4} is positive and x-\frac{4}{3} is negative.

$$x+\frac{1}{4}>0 x-\frac{4}{3}<0$$

The solution satisfying both inequalities is x\in \left(-\frac{1}{4},\frac{4}{3}\right).

$$x\in \left(-\frac{1}{4},\frac{4}{3}\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(-\frac{1}{4},\frac{4}{3}\right)$$