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7n^{2}-121n-3856\geq 0
Subtract 3728 from -128 to get -3856.
7n^{2}-121n-3856=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(-121\right)±\sqrt{\left(-121\right)^{2}-4\times 7\left(-3856\right)}}{2\times 7}
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 7 for a, -121 for b, and -3856 for c in the quadratic formula.
n=\frac{121±\sqrt{122609}}{14}
Do the calculations.
n=\frac{\sqrt{122609}+121}{14} n=\frac{121-\sqrt{122609}}{14}
Solve the equation n=\frac{121±\sqrt{122609}}{14} when ± is plus and when ± is minus.
7\left(n-\frac{\sqrt{122609}+121}{14}\right)\left(n-\frac{121-\sqrt{122609}}{14}\right)\geq 0
Rewrite the inequality by using the obtained solutions.
n-\frac{\sqrt{122609}+121}{14}\leq 0 n-\frac{121-\sqrt{122609}}{14}\leq 0
For the product to be ≥0, n-\frac{\sqrt{122609}+121}{14} and n-\frac{121-\sqrt{122609}}{14} have to be both ≤0 or both ≥0. Consider the case when n-\frac{\sqrt{122609}+121}{14} and n-\frac{121-\sqrt{122609}}{14} are both ≤0.
n\leq \frac{121-\sqrt{122609}}{14}
The solution satisfying both inequalities is n\leq \frac{121-\sqrt{122609}}{14}.
n-\frac{121-\sqrt{122609}}{14}\geq 0 n-\frac{\sqrt{122609}+121}{14}\geq 0
Consider the case when n-\frac{\sqrt{122609}+121}{14} and n-\frac{121-\sqrt{122609}}{14} are both ≥0.
n\geq \frac{\sqrt{122609}+121}{14}
The solution satisfying both inequalities is n\geq \frac{\sqrt{122609}+121}{14}.
n\leq \frac{121-\sqrt{122609}}{14}\text{; }n\geq \frac{\sqrt{122609}+121}{14}
The final solution is the union of the obtained solutions.