Solving quadratic vector equation

Hope it is a right place to ask how to solve the equation on $\mathbf x$: $$ \mathbf x^T \mathbf A\mathbf x + \mathbf x^T \mathbf b + c = 0. $$ where:
$\mathbf x$ is an $n\times 1$ column vector
$\mathbf A$ is an $n\times n$ matrix
$\mathbf b$ is an $n\times 1$ vector
$c$ is a scalar
Thanks

Hint 1: We can assume ${\bf A}$ is symmetric. If it's not, we can replace it by ${\bf B}=({\bf A}+{\bf A}^T)/2$, because ${\bf x}^T {\bf A}{\bf x} = {\bf x}^T {\bf B}{\bf x}$ (check it). (That's why quadratic equations are usually expressed using symmetric matrices; we don't lose generality).

Hint 2: This is the generalization of the (scalar) quadratic $$ax^2 + b x + c = 0$$ Do you know how to solve it (completing the square)? If so, try to generalize the procedure. If not, learn it.

Hint 3: Consider the special case ${\bf x}^T {\bf x} = v$. If $v$ is positive, the solutions lie on a sphere. Now, if ${\bf x^T A x} = v$ , if we can write ${\bf A} = {\bf P \Lambda P^T }$ (we can if A is symmetric), we make a change of variable ${\bf z} = {\bf P^T x}$ (a rotation of axis) and we get the equation of a (hyper)ellipse, if $ v$ is positive and ${\bf \Lambda }$ is diagonal with positive entries.