How do you find delta in an epsilon-delta proof of continuity when the function is nonlinear and can't be factored? So we define $f:\mathbb{R}\to \mathbb{R}$ and $f(x)=x^2+3x-5$, and we're using the epsilon-delta definition to prove that it is continuous at 3.
So I know the general structure of an epsilon-delta proof, but we were never shown how to prove a nonlinear function. Like I have no idea what to choose delta as, and how to even figure out what we should choose delta as. I know we need to let epsilon be an arbitrary positive real number, and plug the equation so that $|x^2+3x-5-3|<\delta$ so that $|x^2+3x-8|<\delta$. I don't think you can factor this, so what do you do in order to choose delta?

Let $\delta$ be the value we want and lets try to find what restrictions we need.
$|x - 3| < \delta$
$3 -\delta < x < 3 + \delta$.  As $\delta$ can be arbitrarily small we can assume will always pick a $\delta \le 3$.
$0\le 9-6\delta + \delta^2 < x^2 < 9 +6\delta + \delta^2$
$(9 - 6\delta + \delta^2) + 3(3-\delta) -5 < x^2 + 3x -5 < (9 + 6\delta + \delta^2) + 3(3+\delta) - 5$
$13 - 9\delta + \delta^2 < x^2 +3x-5 < 13 +9\delta + \delta^2$
$-9\delta + \delta^2 < (x^2 + 3x - 5) - 13 < 9\delta + \delta^2$.
now as $\delta$ can be arbitrarily small we can assume we will always pick a $\delta \le 1$.  And if $0 < \delta \le 1$ then $\delta^2 \le \delta$.
$9\delta + \delta^2 \le 9\delta + \delta = 10\delta$.  And $-9\delta + \delta^2 >-9\delta > -10\delta$.
So
$-10\delta \le -9\delta + \delta^2 < (x^2 + 3x - 5)-13 <9\delta + \delta^2 < 10 \delta$.
SO
$-10\delta  < (x^2 + 3x - 5)-13  < 10 \delta$
$|(x^2 + 3x - 5)-13| < 10\delta$.
$|(x^2 + 3x -5) - (3^3 + 3\cdot 3 - 5)| < 10\delta$
So if we choose a $\delta$ where $\delta \le \frac \epsilon {10}$ and $\delta \le 1$ and $\delta \le 3$ we will be good and proven our result.
So let $\delta = \min(\frac {\epsilon}{10}, 1)$.
