We want to show that $\lim_{x\to -2}(2x^2+5x+3) = 1$ using $\epsilon$-$\delta$. Or rather, you want to find a $\delta$ such that if $0\lt |x-(-2)|\lt\delta$, then $|(2x^2+5x+3)-1|\lt \epsilon$ for $\epsilon=0.04$.
Note that $2x^2+5x+3-1 = 2x^2+5x+2 = (x+2)(2x+1)$. So we want to control both $|x+2|=|x-(-2)|$ and $|2x+1|$. Note that if $|x+2|\lt 1$, then $-3\lt x\lt -1$, so $-6\lt 2x\lt -2$, and $-5\lt 2x+1\lt -1$, so $1\lt |2x+1|\lt 5$.
So we would like $|x+2|$ to be both less than $1$, and also less than $(0.04)/5 = 0.008$. For example, take $\delta=0.005$. If $0\lt |x+2|\lt 0.005$, then $|2x+1|\lt 5$, and we have: $|(2x^2+5x+3)-1| = |(x+2)(2x+1)| = |x+2|\,|2x+1|\lt (0.005)5 = 0.025\lt 0.05=\epsilon$ so this $\delta$ suffices.