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10n^{2}-383n+2079=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(-383\right)±\sqrt{\left(-383\right)^{2}-4\times 10\times 2079}}{2\times 10}
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 10 for a, -383 for b, and 2079 for c in the quadratic formula.
n=\frac{383±\sqrt{63529}}{20}
Do the calculations.
n=\frac{\sqrt{63529}+383}{20} n=\frac{383-\sqrt{63529}}{20}
Solve the equation n=\frac{383±\sqrt{63529}}{20} when ± is plus and when ± is minus.
10\left(n-\frac{\sqrt{63529}+383}{20}\right)\left(n-\frac{383-\sqrt{63529}}{20}\right)\leq 0
Rewrite the inequality by using the obtained solutions.
n-\frac{\sqrt{63529}+383}{20}\geq 0 n-\frac{383-\sqrt{63529}}{20}\leq 0
For the product to be ≤0, one of the values n-\frac{\sqrt{63529}+383}{20} and n-\frac{383-\sqrt{63529}}{20} has to be ≥0 and the other has to be ≤0. Consider the case when n-\frac{\sqrt{63529}+383}{20}\geq 0 and n-\frac{383-\sqrt{63529}}{20}\leq 0.
n\in \emptyset
This is false for any n.
n-\frac{383-\sqrt{63529}}{20}\geq 0 n-\frac{\sqrt{63529}+383}{20}\leq 0
Consider the case when n-\frac{\sqrt{63529}+383}{20}\leq 0 and n-\frac{383-\sqrt{63529}}{20}\geq 0.
n\in \begin{bmatrix}\frac{383-\sqrt{63529}}{20},\frac{\sqrt{63529}+383}{20}\end{bmatrix}
The solution satisfying both inequalities is n\in \left[\frac{383-\sqrt{63529}}{20},\frac{\sqrt{63529}+383}{20}\right].
n\in \begin{bmatrix}\frac{383-\sqrt{63529}}{20},\frac{\sqrt{63529}+383}{20}\end{bmatrix}
The final solution is the union of the obtained solutions.