Solve 27c^2-27c-22>0

$$27 c ^ { 2 } - 27 c - 22 > 0$$

Solve for c

$$c\in \left(-\infty,-\frac{\sqrt{345}}{18}+\frac{1}{2}\right)\cup \left(\frac{\sqrt{345}}{18}+\frac{1}{2},\infty\right)$$

Steps Using the Quadratic Formula

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.

$$27c^{2}-27c-22=0$$

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 27 for a, -27 for b, and -22 for c in the quadratic formula.

$$c=\frac{-\left(-27\right)±\sqrt{\left(-27\right)^{2}-4\times 27\left(-22\right)}}{2\times 27}$$

Do the calculations.

$$c=\frac{27±3\sqrt{345}}{54}$$

Solve the equation c=\frac{27±3\sqrt{345}}{54} when ± is plus and when ± is minus.

$$c=\frac{\sqrt{345}}{18}+\frac{1}{2} c=-\frac{\sqrt{345}}{18}+\frac{1}{2}$$

Rewrite the inequality by using the obtained solutions.

$$27\left(c-\left(\frac{\sqrt{345}}{18}+\frac{1}{2}\right)\right)\left(c-\left(-\frac{\sqrt{345}}{18}+\frac{1}{2}\right)\right)>0$$

For the product to be positive, c-\left(\frac{\sqrt{345}}{18}+\frac{1}{2}\right) and c-\left(-\frac{\sqrt{345}}{18}+\frac{1}{2}\right) have to be both negative or both positive. Consider the case when c-\left(\frac{\sqrt{345}}{18}+\frac{1}{2}\right) and c-\left(-\frac{\sqrt{345}}{18}+\frac{1}{2}\right) are both negative.

$$c-\left(\frac{\sqrt{345}}{18}+\frac{1}{2}\right)<0 c-\left(-\frac{\sqrt{345}}{18}+\frac{1}{2}\right)<0$$

The solution satisfying both inequalities is c<-\frac{\sqrt{345}}{18}+\frac{1}{2}.

$$c<-\frac{\sqrt{345}}{18}+\frac{1}{2}$$

Consider the case when c-\left(\frac{\sqrt{345}}{18}+\frac{1}{2}\right) and c-\left(-\frac{\sqrt{345}}{18}+\frac{1}{2}\right) are both positive.

$$c-\left(-\frac{\sqrt{345}}{18}+\frac{1}{2}\right)>0 c-\left(\frac{\sqrt{345}}{18}+\frac{1}{2}\right)>0$$

The solution satisfying both inequalities is c>\frac{\sqrt{345}}{18}+\frac{1}{2}.

$$c>\frac{\sqrt{345}}{18}+\frac{1}{2}$$

The final solution is the union of the obtained solutions.

$$c<-\frac{\sqrt{345}}{18}+\frac{1}{2}\text{; }c>\frac{\sqrt{345}}{18}+\frac{1}{2}$$