#### Answer

We take $\delta=1/2$ here.

#### Work Step by Step

$$a=-7/2 \hspace{1cm} b=-1/2\hspace{1cm}c=-3$$
Here we need to find $\delta\gt0$ such that for all $x$, $0\lt |x-(-3)|\lt\delta$, or $0\lt|x+3|\lt\delta$, then $-7/2\lt x\lt -1/2$.
The sketch is shown below.
The nearer endpoint of $(-7/2,-1/2)$ to $c=-3$ to $-7/2$, and the distance is $1/2$.
So if we take $\delta=1/2$ or any smaller positive value, then $0\lt |x+3|\lt1/2$, and $x$ will only range in $(-7/2,-5/2)$, as shown by the red lines in the sketch, thereby $x$ is always placed between $-7/2$ and $-1/2$ as requested.
In other words, $$0\lt |x+3|\lt\frac{1}{2}\hspace{2cm}-\frac{7}{2}\lt x\lt -\frac{1}{2}$$