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3x^{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{-5±\sqrt{5^{2}-4\times 3\left(-1\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 -1 for c in the quadratic formula.
x=\frac{-5±\sqrt{37}}{6}
Do the calculations.
x=\frac{\sqrt{37}-5}{6} x=\frac{-\sqrt{37}-5}{6}
Solve the equation x=\frac{-5±\sqrt{37}}{6} when ± is plus and when ± is minus.
3\left(x-\frac{\sqrt{37}-5}{6}\right)\left(x-\frac{-\sqrt{37}-5}{6}\right)\geq 0
Rewrite the inequality by using the obtained solutions.
x-\frac{\sqrt{37}-5}{6}\leq 0 x-\frac{-\sqrt{37}-5}{6}\leq 0
For the product to be ≥0, x-\frac{\sqrt{37}-5}{6} and x-\frac{-\sqrt{37}-5}{6} have to be both ≤0 or both ≥0. Consider the case when x-\frac{\sqrt{37}-5}{6} and x-\frac{-\sqrt{37}-5}{6} are both ≤0.
x\leq \frac{-\sqrt{37}-5}{6}
The solution satisfying both inequalities is x\leq \frac{-\sqrt{37}-5}{6}.
x-\frac{-\sqrt{37}-5}{6}\geq 0 x-\frac{\sqrt{37}-5}{6}\geq 0
Consider the case when x-\frac{\sqrt{37}-5}{6} and x-\frac{-\sqrt{37}-5}{6} are both ≥0.
x\geq \frac{\sqrt{37}-5}{6}
The solution satisfying both inequalities is x\geq \frac{\sqrt{37}-5}{6}.
x\leq \frac{-\sqrt{37}-5}{6}\text{; }x\geq \frac{\sqrt{37}-5}{6}
The final solution is the union of the obtained solutions.