Question: Determine the number and location of the equilibrium points for all values of $c$ of
$\dot{x}=x^2+2x+c+2$, $(1)$
$x \in \mathbb{R}$, where $c \in \mathbb{R}$, constant, is a control parameter for the system.
My Answer: $\dot{x}=0$ for stationary points. So, $x^2+2x+c+2=0$. Using the quadratic formula we get,
$x=\frac{-2± \sqrt{2^2-4(c+2)}}{2}=-1±\sqrt{-1-c}$ .
If $c=-1$: then $x=-1$ is a stationary point of $(1)$.
If $c<-1$: then $x=-1±\sqrt{\alpha}$ where $\alpha=-1-c>0$ are stationary points of $(1)$.
If $c>-1$: then $x=-1±\sqrt{\beta}i$ where $\beta=1+c>0$ are stationary points of $(1)$.
I am not sure if what I have done is correct as I was not sure how to deal with $c$ being unknown. Is this all I needed to do or have I missed some points?
Your approach is indeed correct. You get the stationary points (equilibriums) by finding the solutions to $\dot{x} = 0 \Rightarrow x^2 + 2x + c + 2 = 0$.
Your answer regarding $c=-1$ and $-1-c>0$ is correct, but you have a mistake for the case of $-1-c < 0$. Take note that your variable $x$ is over the reals, $x \in \mathbb R$. This means that this is an ODE over the reals, which means that there is no equilibria for $-1-c <0$, as these will be complex numbers $\notin \mathbb R$.