I have two quantities which came from solving quadratic equations: $\frac{n+1+\sqrt{28n^2+20n-71}}{9}$ and $\frac{n+3+\sqrt{28n^2-12n+9}}{9}$. I can see that in the limit as $n \to \infty$ they approach the same value: $\frac{1+2\sqrt{7}}{9} \cdot n$. These two quantities are in fact very close together, graphically it looks like their difference is less than 1 in absolute value for $n \ge 2$. How can I show that?
Manipulating equations involving square roots is already tricky, but attempting those same techniques here only seems to work at proving the upper bound, namely: $$\frac{n+1+\sqrt{28n^2+20n-71}}{9} - \frac{n+3+\sqrt{28n^2-12n+9}}{9} \le 1$$ When $n \ge$ 1.8ish. How would you show the lower bound of -1 when $n \ge 2$?
Note that \begin{align} \Delta &=\frac{n+1+\sqrt{28n^2+20n-71}}{9} - \frac{n+3+\sqrt{28n^2-12n+9}}{9}\\ & = -\frac{2}{9}+\frac{\sqrt{28n^2+20n-71}-\sqrt{28n^2-12n+9}}{9} \\ & = -\frac{2}{9}+\frac{28n^2+20n-71-28n^2+12n-9}{9(\sqrt{28n^2+20n-71}+\sqrt{28n^2-12n+9})}\\ & = -\frac{2}{9}+\underbrace{\frac{32n-80}{9(\sqrt{28n^2+20n-71}+\sqrt{28n^2-12n+9})}}_{\text{increasing function of } n}\\ & \ge -\frac{2}{9}+{\frac{32(2)-80}{9(\sqrt{28(2)^2+20(2)-71}+\sqrt{28(2)^2-12(2)+9})}}\\ & \ge -\frac{2}{9}-\frac{16}{9(9+\sqrt{97})}\\ & \ge -1. \end{align}