Answer
$(-2.01,-1.99)$
$\delta=0.01$
Work Step by Step
We are given $f(x)=2x-2$, $L=-6$, $c=-2$, and $\epsilon=0.02$.
To find the desired interval about $c$, we plug the above into $|f(x)-L| \lt \epsilon$ and solve for $x$:
$$\left|2x-2-(-6) \right| \lt 0.02\\
\left|2x+4\right| \lt 0.02
-0.02 \lt 2x+4 \lt 0.02\\
-0.01\lt x+2 \lt 0.01\\
-2.01 \lt x \lt -1.99.$$
Thus our interval is $(-2.01,-1.99)$.
Now, to find our $\delta$, we note that
$$-2.01 \lt x \lt -1.99 \implies -0.01 \lt x-(-2) \lt 0.01.$$
Hence for $\delta=0.01$,
$$0 \lt |x-(-2)| \lt \delta \implies |f(x)-(-6)| \lt \epsilon.$$
