We have been challenged with an $\epsilon$, perhaps $\epsilon=1/1000$. We want to come up with a $\delta$ such that if $|x-2|\lt \delta$, then for sure $|x^2+3x-10|\lt \epsilon$.
Suppose that after some calculation, we announce that $\delta=\epsilon/8$ does the job. Then triumphantly the challenger could say that she had $\epsilon=47$ in mind, and that in that case taking $\delta=47/8$ is insufficient. Of course, that is not playing fair. But we might as well come up with a $\delta=\delta(\epsilon)$ that always works.
Some algebra shows that $x^2+3x-10=(x-2)^2+7(x-2)$. So we want to make sure that $|(x-2)^2+7(x-2)|\lt \epsilon$. Note that
$$|(x-2)^2+7(x-2)|\le (x-2)^2+7|x-2|.$$
We want to make the right-hand side "small," by choosing $x$ appropriately close to $2$. Suppose we had been given a ridiculous $\epsilon$, like $47$. If $x$ is within $47$ of $2$, the number $(x-2)^2$ could be very large. So the first task is to make sure $\delta$ is small enough not to allow $(x-2)^2$ to be large.
So we say first of all, let $\delta\le 1$. Then if $|x-2|\lt \delta$, it follows that $(x-2)^2\lt \delta$. For $(x-2)^2=(x-2)(x-2)$. The "first" $x-2$ has absolute value $\le 1$, and the second has absolute value $|x-2|$, so the product has absolute value $\le |x-2|$.
It follows that as long as $\delta\le 1$, we have $(x-2)^2+7|x-2|\le 8|x-2|$. To make sure this is $\lt \epsilon$, it is enough to make $\delta=\frac{1}{8}\epsilon$.
However, in deriving our simplified inequality, we assumed that $\delta\le 1$. So we know that everything will work if $\delta=\min(1,\epsilon/8)$.
It might have been better to observe that
$$(x-2)^2+7|x-2|=|x-2|\left(|x-2|+7\right).$$
Then it is clear that if $|x-2|\le 1$, then $(x-2)^2+7|x-2|\le 8|x-2|$.