Still me. It is possible to solve the quadratic programming with simplex method.
Method 1
The Lagrangian function of original problem is
\begin{equation}
L(x; \lambda_1, \lambda2) = \frac{1}{2}x^TQx + c^Tx + \lambda_1^T(Ax-b) - \lambda_2^T x, \quad \lambda_1, \lambda_2 \ge 0.
\end{equation}
And the KKT conditions are
\begin{equation}
\begin{array}{c}
{\nabla_x L(x;\lambda_1,\lambda_2) = Qx+c+A\lambda_1-\lambda_2 = 0.} \\
{Ax-b \leq 0, \quad x \ge 0.} \\
{\lambda_1 \ge 0, \quad \lambda_2 \ge 0.} \\
{\lambda_1^T(Ax-b)=0, \quad \lambda_2^T x= 0.}
\end{array}
\end{equation}
Except the last complementary slack condition, All of them are linear. We temperately drop this condition, and add one slack variable $y$, then can obtain
\begin{equation}
\begin{array}{c}
{Qx+c+A\lambda_1-\lambda_2 = 0} \\
{Ax-b+y = 0} \\
{x \ge 0, \lambda_1 \ge 0, \lambda_2 \ge 0, y \ge 0}
\end{array}
\end{equation}
And we can use simplex method to search feasible solution of the above constraints.
However, to satisfied the complementary slack condition,
we must take one of ${\lambda_1}_i$ and $y_i$ as zero.
Similarly, we also must take one of ${\lambda_2}_i$ and $x_i$ as zero.
In other words, we need to keep ${\lambda_1}_i$ and $y_i$, ${\lambda_2}_i$ and $x_i$ could not be basis variables at the same time.
Method 2
But now I guess what you want is another method which is called Frank-Wolfe algorithm. For the iterative point $x_k$ at $k$ step, we can compute the gradient
\begin{equation}
\nabla_x J(x_k) = Qx_k + c
\end{equation}
And we try to solve the following approximated problem:
\begin{equation}
\begin{array}{cl}
{\min_{x}} & J(x_k)+ \nabla_{x}J(x_k)^T(x-x_k) \\
{} & {Ax -b \leq 0} \\
{} & {x \ge 0}
\end{array}
\end{equation}
This is a linear programming in terms of $x$, and you can solve it directly by the Simplex method.
##Method 1and##Method 2. Also, regarding the 1st method, your last sentence isn't clear. – Royi Mar 30 '20 at 08:07