Given,

$\left|n^2-10n+19\right|<6$

Now we know that,

$\left|x\right|<a\Rightarrow -a<x<a$

Now using above formula we get,

$\Rightarrow -6<n^2-10n+19<6$

$\Rightarrow n^2-10n+25>0$ and $n^2-10n+13<0$

$\Rightarrow {\left(n-5\right)}^2>0$ and $\left(n-\left(5-2\sqrt{3}\right)\right)\left(n-\left(5+2\sqrt{3}\right)\right)<0$

$\Rightarrow n\in \mathrm{\mathbb{Z}}-\left\{5\right\}$ and $n\in \left\{2, 3, 4, 5, 6, 7, 8\right\}$

Hence, $n=2, 3, 4, 6, 7, 8$