Solve 22/27+c<c^2

$$\frac { 22 } { 27 } + c < c ^ { 2 }$$

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

Subtract c^{2} from both sides.

$$\frac{22}{27}+c-c^{2}<0$$

Multiply the inequality by -1 to make the coefficient of the highest power in \frac{22}{27}+c-c^{2} positive. Since -1 is negative, the inequality direction is changed.

$$-\frac{22}{27}-c+c^{2}>0$$

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.

$$-\frac{22}{27}-c+c^{2}=0$$

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 1 for a, -1 for b, and -\frac{22}{27} for c in the quadratic formula.

$$c=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 1\left(-\frac{22}{27}\right)}}{2}$$

Do the calculations.

$$c=\frac{1±\frac{1}{9}\sqrt{345}}{2}$$

Solve the equation c=\frac{1±\frac{1}{9}\sqrt{345}}{2} 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.

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

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.

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