Answer
$\delta=-\dfrac{2-\sqrt{5}}{2} \gt 0$
Work Step by Step
We see from the graph that in order for $f(x)$ to be within $\epsilon=0.25$ of $L=3$, we must have
$$-\dfrac{\sqrt{5}}{2} \lt x \lt -\dfrac{\sqrt{3}}{2}.$$
Subtracting $c=-1$ from all three sides gives
$$\dfrac{2-\sqrt{5}}{2} \lt x-(-1) \lt \dfrac{2-\sqrt{3}}{2}.$$
Now, $\dfrac{2-\sqrt{5}}{2} \approx -0.118$ and $\dfrac{2-\sqrt{3}}{2} \approx 0.134$.
This means $\left|\dfrac{2-\sqrt{5}}{2}\right| \lt \left|\dfrac{2-\sqrt{3}}{2}\right|$.
Thus,
$$\dfrac{2-\sqrt{5}}{2} \lt x-(-1) \lt -\dfrac{2-\sqrt{5}}{2} \implies \dfrac{2-\sqrt{5}}{2} \lt x-(-1) \lt \dfrac{2-\sqrt{3}}{2}.$$
Hence for $\delta = -\dfrac{2-\sqrt{5}}{2} \gt 0$,
$$0 \lt |x-(-1)| \lt \delta \implies 0 \lt |f(x)-3| \lt \epsilon.$$
