I would write:

\(\displaystyle f(g(x))=\sqrt{4-(2x+3)^2}\)

Now, we require the radicand to be non-negative:

\(\displaystyle 4-(2x+3)^2\ge0\)

Factor the LHS:

\(\displaystyle (2+(2x+3))(2-(2x+3))\ge0\)

Distribute:

\(\displaystyle (2+2x+3)(2-2x-3)\ge0\)

Combine like terms:

\(\displaystyle (2x+5)(-2x-1)\ge0\)

Multiply by -1:

\(\displaystyle (2x+5)(2x+1)\le0\)

Since we have a quadratic on the left, whose graph is a parabola which opens up, we know it will be non-positive between and including the roots, which are:

\(\displaystyle x\in\left\{-\frac{5}{2},-\frac{1}{2}\right\}\)

Hence, the solution is given by:

\(\displaystyle -\frac{5}{2}\le x\le-\frac{1}{2}\)

Does that make sense?