Is this $\epsilon$-$\delta$ limit correct? I have to show that $\lim_{x\to2}x^2+x+1=7$ with the $\epsilon$-$\delta$ definition of limit. Here is how I have done it: $\left\lvert x^2+x-6\right\rvert=\lvert x−2\rvert\lvert x+3\rvert<\epsilon$ and since we are close to $2$, we can assume that the $\delta$-neighborhood of $c=2$ must be have a radius of max $\delta=1$ which implies that : $\lvert x-2\rvert < \frac{\epsilon}{6}$.
We now choose $\delta = \min\{3,\frac{\epsilon}{6}\}$  and we can conclude that if $\lvert x−2\rvert<\delta$, it follows that $\left\lvert x^2+x+1-7\right\rvert < 6\frac{\epsilon}{6}=\epsilon$. Is this calculation correct? Do I miss something? Or some details?

We put $f(x) = x^2 +x+1$
We can prove that :
$|f(x) - l|<\delta $ $ \Rightarrow $ $ |x-a|<\alpha $
$\alpha , \delta > 0$
$|f(x) - l|= |x^2 +x+1-7|=|x^2 +x-6|=|x-2||x+3|$
$|f(x) - l|<\delta$
$\Rightarrow $ $|x-2||x+3|<\delta$
Suppose $x\in [\frac{3}{2}, \frac{5}{2}] $
$\Rightarrow $$\frac{9}{2}\leq x+3\leq\frac{11}{2}$
$\Rightarrow $$|x+3|\leq \frac {11}{2} $
$\Rightarrow $$|x-2||x+3|\leq \frac{11}{2} |x-2|$
we know that :
$|x-2||x+3|<\delta$
So:
$\frac{11}{2} |x-2|<\delta$
$\Rightarrow $ $|x-2|<\frac{2} {11}\delta$
We put $\alpha=\frac{2\delta} {11}$
Finally :
After the definition of limite we proved $\lim_{x\to 2} f(x) =7$
