It doesn't look like. The intuition that for example when $A$ is positive definite you can find a minimum and check if it is less than $c$.
Here is the approach to solve these equations.
Firstly, diagonalize $A = Q \Lambda Q^T$, where $Q$ is unitary. Then you have
$$
x^T Q \Lambda Q^T x + b^Tx = c
$$
Substitute $y = Q^T x$ (then $x = Qy$).
You have
$$
y^T\Lambda y + b^TQy = c
$$
Here, denote $d^T = b^TQ$, and you have very simple equation
$$
y^T \Lambda y + d^Ty = c
$$ where $\Lambda$ is a diagonal matrix of eigenvalues.
You can do even more and get rid of a linear part by taking the square like $ax^2+bx = a(x^2 + \frac{b}{a}x) = a(x+\frac{b}{2a})^2-\frac{b^2}{4a}$ for every coordinate. Here several cases follows:
- if all eigenvalues $>0$ you have paraboloid so simply finding minimum and checking if it is less than $c$ will answer your question.
- if all eigenvalues $<0$ you have paraboloid and checking it maximum and checking if it is greater than $c$ will answer the question.
- if there is an eigenvalue 0 and corresponding linear coefficient is not 0 than solution exists (for some $i$ you have $d_iy_i = c$, then $y_i = \frac{c}{d_i}$).
- You have a saddle i.e. there is a eigenvalue > 0 and eigenvalue < 0. Then tending those coordinates to plus/minus infinity you see that your functions image is the whole $\mathbb{R}$ so there must be a solution.
- if all eigenvalues and linear coefficients are 0 then there is a solution iff $c=0$.
The $\mathcal{NP}$-hard part comes into a play when you have restrictions on the domain for example a unit box.