Solve 12x^2-12^2x+9>0

$$12 { x }^{ 2 } - { 12 }^{ 2 } x+9 > 0$$

Solve for x

$$x\in \left(-\infty,-\frac{\sqrt{141}}{2}+6\right)\cup \left(\frac{\sqrt{141}}{2}+6,\infty\right)$$

Steps Using the Quadratic Formula

Calculate 12 to the power of 2 and get 144.

$$12x^{2}-144x+9>0$$

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.

$$12x^{2}-144x+9=0$$

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 12 for a, -144 for b, and 9 for c in the quadratic formula.

$$x=\frac{-\left(-144\right)±\sqrt{\left(-144\right)^{2}-4\times 12\times 9}}{2\times 12}$$

Do the calculations.

$$x=\frac{144±12\sqrt{141}}{24}$$

Solve the equation x=\frac{144±12\sqrt{141}}{24} when ± is plus and when ± is minus.

$$x=\frac{\sqrt{141}}{2}+6 x=-\frac{\sqrt{141}}{2}+6$$

Rewrite the inequality by using the obtained solutions.

$$12\left(x-\left(\frac{\sqrt{141}}{2}+6\right)\right)\left(x-\left(-\frac{\sqrt{141}}{2}+6\right)\right)>0$$

For the product to be positive, x-\left(\frac{\sqrt{141}}{2}+6\right) and x-\left(-\frac{\sqrt{141}}{2}+6\right) have to be both negative or both positive. Consider the case when x-\left(\frac{\sqrt{141}}{2}+6\right) and x-\left(-\frac{\sqrt{141}}{2}+6\right) are both negative.

$$x-\left(\frac{\sqrt{141}}{2}+6\right)<0 x-\left(-\frac{\sqrt{141}}{2}+6\right)<0$$

The solution satisfying both inequalities is x<-\frac{\sqrt{141}}{2}+6.

$$x<-\frac{\sqrt{141}}{2}+6$$

Consider the case when x-\left(\frac{\sqrt{141}}{2}+6\right) and x-\left(-\frac{\sqrt{141}}{2}+6\right) are both positive.

$$x-\left(-\frac{\sqrt{141}}{2}+6\right)>0 x-\left(\frac{\sqrt{141}}{2}+6\right)>0$$

The solution satisfying both inequalities is x>\frac{\sqrt{141}}{2}+6.

$$x>\frac{\sqrt{141}}{2}+6$$

The final solution is the union of the obtained solutions.

$$x<-\frac{\sqrt{141}}{2}+6\text{; }x>\frac{\sqrt{141}}{2}+6$$