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n^{2}-4019n+4036081\leq 0
Calculate 2009 to the power of 2 and get 4036081.
n^{2}-4019n+4036081=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.
n=\frac{-\left(-4019\right)±\sqrt{\left(-4019\right)^{2}-4\times 1\times 4036081}}{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, -4019 for b, and 4036081 for c in the quadratic formula.
n=\frac{4019±3\sqrt{893}}{2}
Do the calculations.
n=\frac{3\sqrt{893}+4019}{2} n=\frac{4019-3\sqrt{893}}{2}
Solve the equation n=\frac{4019±3\sqrt{893}}{2} when ± is plus and when ± is minus.
\left(n-\frac{3\sqrt{893}+4019}{2}\right)\left(n-\frac{4019-3\sqrt{893}}{2}\right)\leq 0
Rewrite the inequality by using the obtained solutions.
n-\frac{3\sqrt{893}+4019}{2}\geq 0 n-\frac{4019-3\sqrt{893}}{2}\leq 0
For the product to be ≤0, one of the values n-\frac{3\sqrt{893}+4019}{2} and n-\frac{4019-3\sqrt{893}}{2} has to be ≥0 and the other has to be ≤0. Consider the case when n-\frac{3\sqrt{893}+4019}{2}\geq 0 and n-\frac{4019-3\sqrt{893}}{2}\leq 0.
n\in \emptyset
This is false for any n.
n-\frac{4019-3\sqrt{893}}{2}\geq 0 n-\frac{3\sqrt{893}+4019}{2}\leq 0
Consider the case when n-\frac{3\sqrt{893}+4019}{2}\leq 0 and n-\frac{4019-3\sqrt{893}}{2}\geq 0.
n\in \begin{bmatrix}\frac{4019-3\sqrt{893}}{2},\frac{3\sqrt{893}+4019}{2}\end{bmatrix}
The solution satisfying both inequalities is n\in \left[\frac{4019-3\sqrt{893}}{2},\frac{3\sqrt{893}+4019}{2}\right].
n\in \begin{bmatrix}\frac{4019-3\sqrt{893}}{2},\frac{3\sqrt{893}+4019}{2}\end{bmatrix}
The final solution is the union of the obtained solutions.