Thread: help with epsilon delta proof

I've read what's posted in the sticky section but I'm still a little confused on one part.

Here's my problem:

Find the limit L, then find a formula for delta>0 such that abs(f(x)-L)<epsilon whenever 0<abs(x-c)<delta for:

lim(2x^2-3x+1)
x is approaching 4

Here's my work:

The limit is 21

abs(2x^2-3x+1-21)<epsilon

abs(2x^2-3x-20)<epsilon

abs((x-4)(2x+5))<epsilon

ok, now I get lost: abs(x-4)<epsilon/abs(2x+5)

I don't know what to do at this point. Can someone help

There was a similar problem on the pdf. under the sticky section (Example 4) involving (2x^2-x+3)=31

as they do the work they get < 17(epsilon/17), I can't see where they get 17, it seems like an arbitrary number.

It's been a while for me, but as I recall, with a quadratic, you have to restrict the values of x in which you're interested. You are interested in values of x close to 4. Well, then, suppose that 3<x<5. Can you now put a bound on the factor 2x+5? You know that x-4 will get small, because x is going to 4. You just need a bound on the other term. Perhaps you can continue from there?