I have the following equations and I am trying to solve them while taking $\lambda$ $\neq$ $0$.
Eq 1: $4x_1^3$ - $2 \lambda x_1$ = $0$
Eq 2: $-4x_2^2$ - $2 \lambda x_2$ - $\lambda$ - $1$ = $0$
Eq 3: $\lambda ( x_1^2+ x_2^2 + x_2)$ = $0$
I have been trying to solve $\lambda$, $x_1$, and $x_2$ but going around in circles for this case. I found (0,0) and (0, -1) but nothing afterwards.
If you are trying to find the minimum of $f(x_1,x_2)=x_1^2-2 x_2^2-x_2$, subject to $x_1^2+x_2^2+x_2 \leq 0$, the proper KKT system would be $$ \begin{cases} 2x_1 - 2 \lambda x_1 = 0\\ -4 x_2-1 -\lambda(2x_2+1)=0\\ \lambda (x_1^2+x_2^2+x_2) = 0\\ x_1^2+x_2^2+x_2 \leq 0\\ \lambda \leq 0 \end{cases}. $$ How did you get to the system you presented?
Regarding the amended system, you still have a typo in the second equation, it should be: $-4x_2 - 1 -2 \lambda x_2 -\lambda = 0$. Once you solve the system you'll find the following solutions: $(0,0)$, $(0, -\frac 14)$ and $(0,-1)$, all with $\lambda \leq 0$.
The solution $(0,-\frac 14)$ is in fact an interior saddle point and you end up with the two points you had already mentioned in your post. Why were you expecting more minima?