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3x^{2}-5x-6\leq 0
Multiply the inequality by -1 to make the coefficient of the highest power in -3x^{2}+5x+6 positive. Since -1 is negative, the inequality direction is changed.
3x^{2}-5x-6=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 3\left(-6\right)}}{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, -5 for b, and -6 for c in the quadratic formula.
x=\frac{5±\sqrt{97}}{6}
Do the calculations.
x=\frac{\sqrt{97}+5}{6} x=\frac{5-\sqrt{97}}{6}
Solve the equation x=\frac{5±\sqrt{97}}{6} when ± is plus and when ± is minus.
3\left(x-\frac{\sqrt{97}+5}{6}\right)\left(x-\frac{5-\sqrt{97}}{6}\right)\leq 0
Rewrite the inequality by using the obtained solutions.
x-\frac{\sqrt{97}+5}{6}\geq 0 x-\frac{5-\sqrt{97}}{6}\leq 0
For the product to be ≤0, one of the values x-\frac{\sqrt{97}+5}{6} and x-\frac{5-\sqrt{97}}{6} has to be ≥0 and the other has to be ≤0. Consider the case when x-\frac{\sqrt{97}+5}{6}\geq 0 and x-\frac{5-\sqrt{97}}{6}\leq 0.
x\in \emptyset
This is false for any x.
x-\frac{5-\sqrt{97}}{6}\geq 0 x-\frac{\sqrt{97}+5}{6}\leq 0
Consider the case when x-\frac{\sqrt{97}+5}{6}\leq 0 and x-\frac{5-\sqrt{97}}{6}\geq 0.
x\in \begin{bmatrix}\frac{5-\sqrt{97}}{6},\frac{\sqrt{97}+5}{6}\end{bmatrix}
The solution satisfying both inequalities is x\in \left[\frac{5-\sqrt{97}}{6},\frac{\sqrt{97}+5}{6}\right].
x\in \begin{bmatrix}\frac{5-\sqrt{97}}{6},\frac{\sqrt{97}+5}{6}\end{bmatrix}
The final solution is the union of the obtained solutions.