how I should find $\delta$ > 0 that if $|x - 1| < \delta$ then $|\frac{x^2-7}{2x+1} + 2| < 10$?

I got something like that $\frac{|x-1||x + 5|}{|2x+1|} < 10$ and I also proved in previous question that if $|x-1| < 1$ then $|x+5| < 7$ and $2x+1 > 1$.

if $|x-1|<1$ then I know that $|x - 1| < 1 \leq \delta$ then I can say that $\frac{|x-1||x + 5|}{|2x+1|} < 7\delta < 10$ but I don't know how to continue for here...

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Thus to ensure that $|x-1|<1$ follows from $|x-1|<\delta$, you better choose $\delta\le 1$. And to ensure $7\delta<10$ you better ensure $\delta<\frac{10}7$. For both conditions together, it apparently suffices to have $\delta=1$.