Find $\delta$ s.t. $|x-2|<\delta$ means $|x^2+2x-18|<\frac{1}{4}$ I need to find $\delta$ s.t. $|x-2|&lt;\delta$ means $|x^2+2x-18|&lt;\frac{1}{4}$ (as if I was asked to find $\delta$ when finding the limit at 2, with $\varepsilon = \frac{1}{4}$). My issue is that this does not factor nicely, so I am having trouble. When I try to factor out $(x-2)$ I get $x^2+2x-18=(x-2)(x+4)+10$, and this is not helping me to obtain $\delta$. 
Thanks!
hint

$$x^2+2x-8=(x-2)(x+4)$$
$$=(x-2)(x-2+6)$$
$$=(x-2)^2+6(x-2)$$

solve for $\delta$ the inequation

$$\delta^2+6\delta&lt;\frac 14.$$
hint

$$x^2+2x-8=(x-2)(x+4)$$
$$=(x-2)(x-2+6)$$
$$=(x-2)^2+6(x-2)$$

solve for $\delta$ the inequation

$$\delta^2+6\delta&lt;\frac 14.$$