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-6x^{2}+5x-1\geq 0
Swap sides so that all variable terms are on the left hand side. This changes the sign direction.
6x^{2}-5x+1\leq 0
Multiply the inequality by -1 to make the coefficient of the highest power in -6x^{2}+5x-1 positive. Since -1 is negative, the inequality direction is changed.
6x^{2}-5x+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(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 6\times 1}}{2\times 6}
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 6 for a, -5 for b, and 1 for c in the quadratic formula.
x=\frac{5±1}{12}
Do the calculations.
x=\frac{1}{2} x=\frac{1}{3}
Solve the equation x=\frac{5±1}{12} when ± is plus and when ± is minus.
6\left(x-\frac{1}{2}\right)\left(x-\frac{1}{3}\right)\leq 0
Rewrite the inequality by using the obtained solutions.
x-\frac{1}{2}\geq 0 x-\frac{1}{3}\leq 0
For the product to be ≤0, one of the values x-\frac{1}{2} and x-\frac{1}{3} has to be ≥0 and the other has to be ≤0. Consider the case when x-\frac{1}{2}\geq 0 and x-\frac{1}{3}\leq 0.
x\in \emptyset
This is false for any x.
x-\frac{1}{3}\geq 0 x-\frac{1}{2}\leq 0
Consider the case when x-\frac{1}{2}\leq 0 and x-\frac{1}{3}\geq 0.
x\in \begin{bmatrix}\frac{1}{3},\frac{1}{2}\end{bmatrix}
The solution satisfying both inequalities is x\in \left[\frac{1}{3},\frac{1}{2}\right].
x\in \begin{bmatrix}\frac{1}{3},\frac{1}{2}\end{bmatrix}
The final solution is the union of the obtained solutions.