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