Solve 63x^2/64-3x/8-9/4>0

$$\frac { 63 x ^ { 2 } } { 64 } - \frac { 3 x } { 8 } - \frac { 9 } { 4 } > 0$$

Solve for x

$$x\in \left(-\infty,-\frac{4}{3}\right)\cup \left(\frac{12}{7},\infty\right)$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 64, the least common multiple of 64,8,4. Since 64 is positive, the inequality direction remains the same.

$$63x^{2}-8\times 3x-144>0$$

Multiply -8 and 3 to get -24.

$$63x^{2}-24x-144>0$$

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.

$$63x^{2}-24x-144=0$$

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 63 for a, -24 for b, and -144 for c in the quadratic formula.

$$x=\frac{-\left(-24\right)±\sqrt{\left(-24\right)^{2}-4\times 63\left(-144\right)}}{2\times 63}$$

Do the calculations.

$$x=\frac{24±192}{126}$$

Solve the equation x=\frac{24±192}{126} when ± is plus and when ± is minus.

$$x=\frac{12}{7} x=-\frac{4}{3}$$

Rewrite the inequality by using the obtained solutions.

$$63\left(x-\frac{12}{7}\right)\left(x+\frac{4}{3}\right)>0$$

For the product to be positive, x-\frac{12}{7} and x+\frac{4}{3} have to be both negative or both positive. Consider the case when x-\frac{12}{7} and x+\frac{4}{3} are both negative.

$$x-\frac{12}{7}<0 x+\frac{4}{3}<0$$

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

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

Consider the case when x-\frac{12}{7} and x+\frac{4}{3} are both positive.

$$x+\frac{4}{3}>0 x-\frac{12}{7}>0$$

The solution satisfying both inequalities is x>\frac{12}{7}.

$$x>\frac{12}{7}$$

The final solution is the union of the obtained solutions.

$$x<-\frac{4}{3}\text{; }x>\frac{12}{7}$$