Given a set of multivariate, quadratic, non-homogeneous equations, is there a way to decide how many non-negative roots it can have?

Some explanations:

1. All the coefficients are real numbers.
2. The number of variables is the same as the number of equations, and all the equations are independent.
3. Some of the equations might actually be linear equations.
4. By "non-negative" solution, I mean a solution in which all the variables take non-negative values.
5. By some "physical" reasoning, we know it must have at least one non-negative solution.

Further explanations:

1. I know generally a set of n quadratic equations with n variables has at most $2^n$ distinct roots.
2. The background of this problem is: the set of quadratic equations is the right-hand side of the chemical rate equation. By equating is to zero, the steady-state case is being considered. As we only take one-body or two-body reactions into account, the degree of the equations are at most two. As the abundance of the molecules cannot be negative, we only care about the non-negative solutions.
3. The number of variables can be up to 1000, so simple numerical test is not practical.

I am not a math student, and I am not sure whether this kind of question is allowed here.

Given a set of multivariate, quadratic, non-homogeneous equations, is there a way to decide how many non-negative roots it have?

Some explanations:

1. All the coefficients are real numbers.
2. The number of variables is the same as the number of equations, and all the equations are independent.
3. Some of the equations might actually be linear equations.
4. By "non-negative" solution, I mean a solution in which all the variables take non-negative values.
5. By some "physical" reasoning, we know it must have at least one non-negative solution.

Further explanations:

1. I know generally a set of n quadratic equations with n variables has at most $2^n$ distinct roots.
2. The background of this problem is: the set of quadratic equations is the right-hand side of the chemical rate equation. By equating is to zero, the steady-state case is being considered. As we only take one-body or two-body reactions into account, the degree of the equations are at most two. As the abundance of the molecules cannot be negative, we only care about the non-negative solutions.
3. The number of variables can be up to 1000, so simple numerical test is not practical.

I am not a math student, and I am not sure whether this kind of question is allowed here.