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