Evaluate the following infinite sum using a similar convergent one $$\sum_{n=1}^\infty \frac{3n^2-4n+2}{n!}$$

I have tried rewriting: $$\frac{3n^2-4n+2}{n!} = \frac{2}{n!} - \frac{1}{(n-1)!} + \frac{3}{(n-2)!}$$

And using: $$\sum_{n=0}^\infty \frac{1}{n!} = e$$
I am also aware of the Shift Rule, but I'm not sure how to apply it here where $N$ is a negative number. I can't seem to put these all together since all my calculations get me nowhere.

Will appreciate any help/hints calculating the second and third term of the sum!

Edit: I've also tried index shifting on the second and third sums, but I end up with the sum starting at $-1$ and I'm unsure how to proceed with a factorial.
As the identity only works as of $n=2$, you can handle the first term separately.

$$\sum_{n=1}^\infty \frac{3n^2-4n+2}{n!}=1+\sum_{n=2}^\infty \frac{3n^2-4n+2}{n!}
\\=1+\sum_{n=2}^\infty \frac{2}{n!} - \sum_{n=2}^\infty \frac{1}{(n-1)!} + \sum_{n=2}^\infty \frac{3}{(n-2)!}
\\=1+\sum_{n=2}^\infty \frac{2}{n!} - \sum_{n=1}^\infty \frac{1}{n!} + \sum_{n=0}^\infty \frac{3}{n!}
\\=1+2(e-2)-(e-1)+3e.
$$
Attention please
$$ S=\sum_{k=1}^{\infty}\frac{3n^2-4n+2}{n!}= \sum_{n=1}^{\infty} \frac{3n(n-1)-n+2}{n!}=$$
$$ \Rightarrow \sum_{n=1}^{\infty} \left( \frac{3}{(n-2)!}- \frac{1}{(n-1)!} +\frac{2}{n!} \right)=\sum_{p=0}^{\infty} \frac{3}{p!}-\sum_{q=0}^{\infty} \frac{1}{q!}+ \sum_{n=1}^{\infty} \frac{2}{n!}= (3e)-(e)+2(e-1)$$ $$\Rightarrow S =4e-2.$$Here we have used $p=n-2, q=n-1$,

Please note that if $N$ is zero, a natural number or non-negative integer, $(-N)!=\pm \infty$, so it does not matter if you get $1/(-2)!$ it will be 0. That is why $p=-1$ is not cosidered as $(-1)!= \pm \infty$ and its reciprocal contributes 0.