How to prove that $ \frac{n+1}{4n^2+3}$ is a Cauchy sequence?

By definition, I need to show that for any $\epsilon \gt 0$ there exists $N \in \mathbb{N}$ such that for any $m,n \gt N$:

$ \lvert a_n - a_m \rvert \lt \epsilon$

So I write,

$ |\frac{n+1}{4n^2+3} - \frac{m+1}{4m^2+3} |\leq | \frac{n+1}{4n^2+3}| + |\frac{m+1}{4m^2+3}| = \frac{n+1}{4n^2+3} + \frac{m+1}{4m^2+3} \lt \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$

From here I try to show,

$\frac{n+1}{4n^2+3} \lt \frac{\epsilon}{2} \quad \quad \Rightarrow \quad \quad 4\epsilon n^2 -2n +3\epsilon -2 \gt 0$

$n \gt \frac{1}{4\epsilon} + \frac{1}{4\epsilon} \sqrt{1-12\epsilon^2+8\epsilon}$

Can someone please explain to me what should I do from here? Am I in the right direction at all?

It's easier to use the fact that, since $n+1\leqslant2n$ and $4n^2+3\geqslant4n^2$, then$$\frac{n+1}{4n^2+3}\leqslant\frac{2n}{4n^2}=\frac1{2n}.$$Now, use the fact that$$\frac1{2n}<\frac\varepsilon2\iff n>\frac1\varepsilon.$$