Solve 3x^2-6x+1>0

$$3 x ^ { 2 } - 6 x + 1 > 0$$

Solve for x

$$x\in \left(-\infty,-\frac{\sqrt{6}}{3}+1\right)\cup \left(\frac{\sqrt{6}}{3}+1,\infty\right)$$

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.

$$3x^{2}-6x+1=0$$

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 3 for a, -6 for b, and 1 for c in the quadratic formula.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 3\times 1}}{2\times 3}$$

Do the calculations.

$$x=\frac{6±2\sqrt{6}}{6}$$

Solve the equation x=\frac{6±2\sqrt{6}}{6} when ± is plus and when ± is minus.

$$x=\frac{\sqrt{6}}{3}+1 x=-\frac{\sqrt{6}}{3}+1$$

Rewrite the inequality by using the obtained solutions.

$$3\left(x-\left(\frac{\sqrt{6}}{3}+1\right)\right)\left(x-\left(-\frac{\sqrt{6}}{3}+1\right)\right)>0$$

For the product to be positive, x-\left(\frac{\sqrt{6}}{3}+1\right) and x-\left(-\frac{\sqrt{6}}{3}+1\right) have to be both negative or both positive. Consider the case when x-\left(\frac{\sqrt{6}}{3}+1\right) and x-\left(-\frac{\sqrt{6}}{3}+1\right) are both negative.

$$x-\left(\frac{\sqrt{6}}{3}+1\right)<0 x-\left(-\frac{\sqrt{6}}{3}+1\right)<0$$

The solution satisfying both inequalities is x<-\frac{\sqrt{6}}{3}+1.

$$x<-\frac{\sqrt{6}}{3}+1$$

Consider the case when x-\left(\frac{\sqrt{6}}{3}+1\right) and x-\left(-\frac{\sqrt{6}}{3}+1\right) are both positive.

$$x-\left(-\frac{\sqrt{6}}{3}+1\right)>0 x-\left(\frac{\sqrt{6}}{3}+1\right)>0$$

The solution satisfying both inequalities is x>\frac{\sqrt{6}}{3}+1.

$$x>\frac{\sqrt{6}}{3}+1$$

The final solution is the union of the obtained solutions.

$$x<-\frac{\sqrt{6}}{3}+1\text{; }x>\frac{\sqrt{6}}{3}+1$$