Consider the following quadratic inequality:
$$x^T Q x + c^T x \leq b$$
for $x\in \Re^{d}$. When does this system have a solution?
Without loss of generality assume that $Q$ is symmetric.
Looking for a complete answer for all cases where $Q$ is 1) positive definite, 2) positive semi-definite, 3) negative definite, 4) negative semi-definite, and 5) indefinite.
It is a useful question with a nice application in quadratically-constrained optimization problems, in which the feasible space is defined by a set of constraints such as the one given above. The answer helps to quickly determine whether a system of quadratic constraints has any feasible solution or not in the preprocessing phase. In fact, if any of the quadratic constraints has no solution, the solver informs that the problem is infeasible before starting the main optimization procedure.
I suppose that $Q$ is definite positive and invertible matrix.
As $Q$ is definite positive, we can write $Q = A^TA$ (the matrix $A \in \mathbb{R}^{n\times n}$ is the square root of the matrix $Q$).
As $Q$ is invertible, $A$ is also invertible matrix.
Then: $$\begin{align} &\iff (Ax)^T(Ax) + c^T x \le b\\ &\iff (Ax)^T(Ax) + \left(\frac{1}{2}\cdot c^TA^{-1}\right) Ax + (Ax)^T \cdot \left( \frac{1}{2}\cdot c^TA^{-1}\right) \le b\\ &\iff (Ax)^T(Ax) + \left(\frac{1}{2}\cdot c^TA^{-1}\right) Ax + (Ax)^T \cdot \left( \frac{1}{2}\cdot A^{-T}c\right) + \frac{1}{4}\cdot c^TA^{-1}\cdot A^{-T}c \le b+ \frac{1}{4}\cdot c^TA^{-1}\cdot A^{-T}c\\ &\iff \left( Ax+ \frac{1}{2}\cdot A^{-T}c\right)^T \left( Ax+ \frac{1}{2}\cdot A^{-T}c\right) \le b+\frac{1}{4}\cdot c^TQ^{-1}c \end{align}$$
If $b+\frac{1}{4}\cdot c^TQ^{-1}c<0$, the problem has no solution.
If $b+\frac{1}{4}\cdot c^TQ^{-1}c \ge 0$, let us make an affine transformation by denoting $$y = Ax+ \frac{1}{2}\cdot A^{-T}c \in \mathbb{R}^n$$
Then, $$\iff y^Ty \le b+\frac{1}{4}\cdot c^TQ^{-1}c \iff \sum_{i=1}^n y_i^2 \le b+\frac{1}{4}\cdot c^TQ^{-1}c$$ We have infinite solutions $y \in \mathbb{R}^n$ inside a n-hypersphere centered at the origine $O$ with radius $\sqrt{b+\frac{1}{4}\cdot c^TQ^{-1}c}$. These solutions $x\in \mathbb{R}^n$ can be found by the inverse of the affine transformation $$x = A^{-1} \left(y- \frac{1}{2}\cdot A^{-T}c\right) = A^{-1}y - \frac{1}{2}\cdot Q^{-1}c$$