From the definition of epsilon delta we know,
$0 < |x-3| < \delta$
And,
$|x^2 + 3x + 2 - 20| < \epsilon$
Or, $|(x + 6)*(x-3)| < \epsilon$ [After factoring out]
Now I am going to go stray from traditional way,
As we don't have any control over $x+6$ let's bound it, we know :
$-\delta < x - 3 < \delta$ or, $9 - \delta < x + 6 < 9 + \delta$
If this is true then this also must be true: $|(x-3)|*|(x+6)| < |(x-3)|*(9+\delta)$
Now we can write it as like this : $|(x-3)|*(9+\delta) < \epsilon$
or, $|(x-3)| < \epsilon/(9+\delta)$
So, $\delta = \epsilon/(9+\delta)$ Using the quadric equation,
$\delta = (-9 + √(81 + 4\epsilon) )/2$
So instead of assuming delta as $1$ or $2$ I just assumed it to be any number and it worked out nicely and tried some values with it and it worked too. So is this correct way of doing it? I know I didn't complete the proof but just wanted to know whether this way is correct or not.
Thank you very much!
Your method seems a bit overcomplicated. Do $$|(x+6)(x-3)|=|(x-3+9)(x-3)|\leq|x-3|^2+9|x-3|$$
We would need $\delta=\min\{1,\varepsilon/10\}$.
Then if $|x-3|<\delta$, we have $|x-3|^2\leq|x-3|$ which means $$|x-3|^2+9|x-3|\leq |x-3|+9|x-3|=10|x-3|$$ and the result follows.