Rewritting the problem:
$$
\mathscr{P}: \min x_1+x_2, \\(2x_1+x_2)^2-4=0, \\-x_1\le0,\\ -x_2\le0
$$

Then the KKT conditions are:
$$
L=(x_1+x_2)+\mu_1(-x_1)+\mu_2(-x_2)+\lambda_1((2x_1+x_2)^2-4)
$$
Stationarity:
$$
\nabla_{x_1}L=1-\mu_1 +\lambda_14(2x_1+x_2)=0\\
\nabla_{x_2}L=1 -\mu_2 +\lambda_12(2x_1+x_2)=0
$$
Feasibility (this is normally redundant and serves only to expose $\nabla L$)
$$
\nabla_{\mu_1}L=-x_1\le0\\
\nabla_{\mu_2}L=-x_2\le0\\
\nabla_{\lambda_1}L=(2x_1+x_2)^2-4=0
$$
Slackness:
$$
\mu_1x_1=0\\
\mu_2x_2=0
$$
The specified point of the full problem is: $[x_1 \ x_2 \ \mu_1 \ \mu_2 \ \lambda_1]=[1 \ 0 \ 0 \ 1/2 \ -1/8]$.

The tangent cone for the solution is evaluated for all the **active** constraints (Mangasarian-Fromovitz constraint qualification), which are only the active inequalities plus the equalities (which will always be active!):
$$
\nabla g_1 \cdot d = \le 0\\
\nabla h_1 \cdot d = 0\\
$$
The condition in this form is trivially met for the given point.

The full Hessian of the **active** constraints must be semi-positive $d'Ld>0$ for $\nabla g_1 d=0$ and $\nabla h_1 d=0$, the orthogonal directions of the **active** constraints.

Hence in the full Hessian:
$$
\nabla^2 L=
\left[
\begin{array}{ccccc}
8\lambda_1 &4\lambda_1 &-1 &0 &4(2x_1+x_2)\\
4\lambda_1 &2\lambda_1 &0 &-1 &2(2x_1+x_2)\\
-1 &0 &0 &0 &0\\
0 &-1 &0 &0 &0\\
4(2x_1+x_2) &2(2x_1+x_2) &0 &0 &0
\end{array}
\right]
=
\left[
\begin{array}{c}
-1 &-2 &-1 &0 &8\\
-1/2 &-1/4 &0 &-1 &4\\
-1 &0 &0 &0 &0\\
0 &-1 &0 &0 &0\\
8 &4 &0 &0 &0
\end{array}
\right]
$$

we discard the 3rd row and column for considering the direction $[1 \ 2 \ 0 \ 0 \ 0]$ for the $h_1$ constraint active, and the rest inactive, and we have the reduced hessian, which **is** positive definite:
$$
L(d)=d'[-1 2][-1 -2;-1/2 -1/4][-1;2]d=3d^2
$$

Note that we cannot choose any other direction without breaking feasibility. So the critical point is a minimum.