I am trying to find the minimum of $-x_1$ with restrictions $\bar g\leq\bar 0$ so that
$$\bar g=\begin{pmatrix} (x_1+2)^2+(x_2-4)^2-20\\ (x_1+2)^2+x_2^2-20\\ -x_1\end{pmatrix}\leq \begin{pmatrix}0\\0\\0\\\end{pmatrix}=\bar0$$
I used KKT conditions to solve this puzzle below so $x_1=\frac{4\mu_1+4\mu_2-1-\mu_3}{2(\mu_1+\mu_2)}$ and $x_2=\frac{4\mu_1}{\mu_1+\mu_2}$ where $\mu_i\in\mathbb R\forall i$. I know from the graphical plot that the solution is something like $(1.5,1.5)$ but I cannot see how I can get such a solution from the equations for $x_1$ and $x_2$.
I followed this part of Wikipedia here, source here, about necessary conditions but I am stack how to find the minimum now. How to find it now with the necessary equations for the optimal point $(x_1,x_2)$?
My calculations
Updates
Wok suggested complementary slackness -assumption $\mu_i g_i(x^*)=0, i=1,2,3$ and dual -feasibility assumption $\mu_i\geq 0,i=1,2,3$. I cannot yet see what it helps here.
I can solve the intersection point just by solving the equations, proof here, but I cannot see how the KKT -way really make a difference in comparison to solving in the easy way, really puzzled!
KKT is some sort of generalization of Lagrangean, example here, trying to understand what is happening...

Try to use primal and dual feasibility and complementary slackness, while assuming $\mu_1+\mu_2\neq 0$ (otherwise your computations of $x_1$ and $x_2$ do not stand). With dual feasibility, at least one of $\mu_1$ and $\mu_2$ are strictly greater than zero.
Finding the solution is a matter of enumerating active constraints.
First case
If $\mu_1=0$ and $\mu_2\neq0$, then $x_2=0$. If $\mu_1\neq0$ and $\mu_2=0$, then $x_2=4$. In both cases, $(x_1+2)^2\leq 4$ (primal feasibility).
Since $x_1\geq0$, $x$ lies on the vertical axis (primal feasibility).
Second case
If $\mu_1\neq0$ and $\mu_2\neq0$, then $x$ is one of the two points at the intersection of the two circles (complementary slackness).
Since $x_1\geq0$, $x$ is the point in the right quadrant. (primal feasibility)
Conclusion
The maximum of $x_1$ is attained in the second case, since $x_1=0$ in the first case.
The solution is $x=(2,2)$.