algorithm to find optimal vertices in convex optimization Given $$\min x^TQx \\ \text {s.t} Ax=b, x\ {>=}\ 0$$
where $Q \not = 0$ and $Q$ is positive semidefinite.
Is there an algorithm that guarantees a vertex solution or at least prefers a vertex solution if exists. I observed that a version of Frank Wolfe's Algorithms with simplex might be useful but there is no proof of that. I need a suggestion of an algorithm that does this (however, not brute force). The answer below doesn't consider the non negative constraint on $x$. solution $y$ is guaranteed to have an optimal vertex solution(s).

Edit: the vertices are on real numbers. This isn't integer programming, but might be helpful in integer programming where $A$ is totally unimodular, but it's LCP form isn't even if for all element $e \in Q$ is in $\{-1,0,1\}$.
 A: $
\def\l{\lambda}
\def\LR#1{\left(#1\right)}
\def\BR#1{\Big(#1\Big)}
\def\trace#1{\operatorname{Tr}\LR{#1}}
\def\qiq{\quad\implies\quad}
\def\p{\partial}
\def\grad#1#2{\frac{\p #1}{\p #2}}
\def\a#1{\color{green}{#1}}
\def\b#1{\color{blue}{#1}}
\def\c#1{\color{red}{#1}}
\def\CLR#1{\c{\LR{#1}}}
\def\fracLR#1#2{\LR{\frac{#1}{#2}}}
$The beauty of a linear constraint is that it can be explicitly solved
$$\eqalign{
Ax = b \qiq
x &= \b{A^+b} \;+\; \CLR{I-A^+A}u \\
  &= \b{x_0} \;+\; \c{P}u \\
}$$
where $u$ is a new unconstrained vector variable, $A^+$ is the pseudoinverse,
and $P$ is an orthoprojector into the nullspace of $A$.
Substituting this into the objective function yields
$$\eqalign{
\l &= x^TQx \\
 &= {\LR{Pu+x_0}^TQ\LR{Pu+x_0}} \\
 &= u^TPQPu + 2x_0^TQPu + x_0^TQx_0 \\
}$$
Calculate the gradient wrt $u$
$$\eqalign{
d\l &= \LR{2u^TPQP + 2x_0^TQP}\:du \\
 &= 2\LR{PQPu + PQx_0}^T\:du \\
\grad{\l}{u} &= 2\LR{PQPu + PQx_0} \\
}$$
Set the gradient to zero and solve for the optimal vector
$$\eqalign{
(PQP)u &= -PQx_0 \\
u &= -\LR{PQP}^+PQx_0 \\
x &= x_0 \;+\; Pu \\
 &= x_0 \;-\; P\LR{PQP}^+PQx_0 \\
 &= \BR{I \,-\, P\LR{PQP}^+PQ}\,A^+b \\
}$$
