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3x^{2}-4x+1=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.
x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 3\times 1}}{2\times 3}
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, -4 for b, and 1 for c in the quadratic formula.
x=\frac{4±2}{6}
Do the calculations.
x=1 x=\frac{1}{3}
Solve the equation x=\frac{4±2}{6} when ± is plus and when ± is minus.
3\left(x-1\right)\left(x-\frac{1}{3}\right)<0
Rewrite the inequality by using the obtained solutions.
x-1>0 x-\frac{1}{3}<0
For the product to be negative, x-1 and x-\frac{1}{3} have to be of the opposite signs. Consider the case when x-1 is positive and x-\frac{1}{3} is negative.
x\in \emptyset
This is false for any x.
x-\frac{1}{3}>0 x-1<0
Consider the case when x-\frac{1}{3} is positive and x-1 is negative.
x\in \left(\frac{1}{3},1\right)
The solution satisfying both inequalities is x\in \left(\frac{1}{3},1\right).
x\in \left(\frac{1}{3},1\right)
The final solution is the union of the obtained solutions.