The number of elements in the set \ n ∈ ℤ: | n ^ 2 - 10n + 19| < 6\...

Given,

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

Now we know that,

\left|x\right| < a ⇒ - a < x < a$</p><p>Now using above formula we get,</p><p>$⇒ - 6 < {n}^{2} - 10n + 19 < 6$</p><p>$⇒ {n}^{2} - 10n + 25 > 0 and {n}^{2} - 10n + 13 < 0$</p><p>$⇒ {\left(n - 5\right)}^{2} > 0$and \left(n - \left(5 - 2\sqrt{3}\right)\right)\left(n - \left(5 + 2\sqrt{3}\right)\right) < 0$

$⇒ n \in \mathbb{Z} - \left\{5\right\}$ and $n \in\left\{2, 3, 4, 5, 6, 7, 8\right\}</p><p>Hence,$n = 2, 3, 4, 6, 7, 8$</p></blockquote>$