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x^{2}-3x-2=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(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 1\left(-2\right)}}{2}
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 1 for a, -3 for b, and -2 for c in the quadratic formula.
x=\frac{3±\sqrt{17}}{2}
Do the calculations.
x=\frac{\sqrt{17}+3}{2} x=\frac{3-\sqrt{17}}{2}
Solve the equation x=\frac{3±\sqrt{17}}{2} when ± is plus and when ± is minus.
\left(x-\frac{\sqrt{17}+3}{2}\right)\left(x-\frac{3-\sqrt{17}}{2}\right)<0
Rewrite the inequality by using the obtained solutions.
x-\frac{\sqrt{17}+3}{2}>0 x-\frac{3-\sqrt{17}}{2}<0
For the product to be negative, x-\frac{\sqrt{17}+3}{2} and x-\frac{3-\sqrt{17}}{2} have to be of the opposite signs. Consider the case when x-\frac{\sqrt{17}+3}{2} is positive and x-\frac{3-\sqrt{17}}{2} is negative.
x\in \emptyset
This is false for any x.
x-\frac{3-\sqrt{17}}{2}>0 x-\frac{\sqrt{17}+3}{2}<0
Consider the case when x-\frac{3-\sqrt{17}}{2} is positive and x-\frac{\sqrt{17}+3}{2} is negative.
x\in \left(\frac{3-\sqrt{17}}{2},\frac{\sqrt{17}+3}{2}\right)
The solution satisfying both inequalities is x\in \left(\frac{3-\sqrt{17}}{2},\frac{\sqrt{17}+3}{2}\right).
x\in \left(\frac{3-\sqrt{17}}{2},\frac{\sqrt{17}+3}{2}\right)
The final solution is the union of the obtained solutions.