The best w_j confusion in xgboost from XGBoost tutorial, it described:

In this equation $w_j$ are independent with respect to each other, the form $G_j w_j + \frac{1}{2}(H_j+λ)w_j^2$ is quadratic and the best $w_j$ for a given structure $q(x)$
and the best objective reduction we can get is:
$w^∗_j = \frac{−G_j}{H_j+λ}$
$obj^∗=\frac{−1}{2}\sum_{j=1}^T\frac{G^2_j}{H_j+λ} + γT$

So my confusion is:
if we would like to minimize the $obj$ function, clearly, the best way is to set:
$G_j w_j + \frac{1}{2}(H_j + λ) w_j^2 = 0$
thus we would have the best $w_j^*$ is:
$w_j^* = \frac{-2G_j}{H_j + λ}$
which is different with the official explanation.
anyone please help to point out where am I wrong?
Usually to find an optimum you set the derivative of the function equal to 0. In your case that gives

$$ \frac{dG_j}{dw_J} = G_j + (H_j+\lambda)w_j = 0 $$

leading to 

$$ w_j = \frac{-G_j}{H_j+\lambda}$$