Solve 28(x^2-1)>33x

$$28 ( x ^ { 2 } - 1 ) > 33 x$$

Solve for x

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

Steps Using the Quadratic Formula

Use the distributive property to multiply 28 by x^{2}-1.

$$28x^{2}-28>33x$$

Subtract 33x from both sides.

$$28x^{2}-28-33x>0$$

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.

$$28x^{2}-28-33x=0$$

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 28 for a, -33 for b, and -28 for c in the quadratic formula.

$$x=\frac{-\left(-33\right)±\sqrt{\left(-33\right)^{2}-4\times 28\left(-28\right)}}{2\times 28}$$

Do the calculations.

$$x=\frac{33±65}{56}$$

Solve the equation x=\frac{33±65}{56} when ± is plus and when ± is minus.

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

Rewrite the inequality by using the obtained solutions.

$$28\left(x-\frac{7}{4}\right)\left(x+\frac{4}{7}\right)>0$$

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

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

The solution satisfying both inequalities is x<-\frac{4}{7}.

$$x<-\frac{4}{7}$$

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

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

The solution satisfying both inequalities is x>\frac{7}{4}.

$$x>\frac{7}{4}$$

The final solution is the union of the obtained solutions.

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