Determining a quadric surface. I've been having great difficulty in determining the type of surface that this thing is:  
$9x^2 - 16xy - 5y^2 + 16xz + 23z^2 = 20x$.
Thoughts.
So, if we took the RHS to be zero, then I know that we could write a symmetric matrix of the quadratic form and find an orthonormal basis such that the matrix of the quadratic form is diagonal. To do this I find that the eigenvalues of the matrix of the quadratic form are $9,-9$ and $27$ and proceed to find orthogonal eigenvectors with the corresponding change of basis matrix e.t.c. so the main diagonal of the diagonal matrix consists of the eigenvalues $-9, 9, 27.$ 
I was wondering, does working all of this out help when it comes to the additional term of $20x$? I.e by finding an orthonormal basis of the quadratic form $9x^2 - 16xy - 5y^2 + 16xz + 23z^2 = 0$ can I quickly determine the shape of the surface in question?
If not, my method would then be to proceed to complete the square on individual terms but I begin to obtain horrific coefficients such as $-\frac{109}{9}$ and $\frac{128}{9}$ and things do not seem to simplify. 
Any guidance will truly help, thank you.
 A: Let us set
$$ \mathbf{x} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}. $$
The homogeneous equation
$$ 9x^2 - 16xy - 5y^2 + 16xz + 23z^2 = 0 $$
without the $20x$ term can be written as $\mathbf{x}^T A \mathbf{x} = 0$ where $A$ is the symmetric matrix
$$ \begin{pmatrix} 9 & -8 & 8 \\ -8 & -5 & 0 \\ -8 & 0 & 23 \end{pmatrix}. $$
Like you noted, this matrix has signature $+,+,-$ so the surface is a double cone. Now you add a linear term to get the equation
$$ 9x^2 - 16xy - 5y^2 + 16xz + 23z^2 - 20x = 0. $$
We might hope that this equation is of the form $(\mathbf{x} - \mathbf{v})^T A (\mathbf{x} - \mathbf{v}) = c$ for some vector $\mathbf{v} = (v_1,v_2,v_3)^T$ because then the surface solving the equation will be a translation of the surface $\mathbf{x}^T A \mathbf{x} = c$ and since we know the signature of $A$, we are done. Note that
$$ c = (\mathbf{x} - \mathbf{v})^T A (\mathbf{x} - \mathbf{v}) = \mathbf{x}^T A \mathbf{x} - 2 \mathbf{v}^T A \mathbf{x} + \mathbf{v}^T A \mathbf{v} \iff \\
\mathbf{x}^T A \mathbf{x} - 2 \mathbf{v}^T A \mathbf{x} = c - \mathbf{v}^T A \mathbf{v}.$$
The term $\mathbf{x}^T A \mathbf{x}$ is the quadratic part and the term $-2 \mathbf{v}^T A \mathbf{x}$ is the linear part. In our case, we want
$$ -2 \mathbf{v}^T A \mathbf{x} = (-20, 0, 0) \mathbf{x} \iff \mathbf{v}^T A \mathbf{x} = (10, 0, 0) \mathbf{x} \iff \mathbf{v}^T A = (10,0,0). $$
Since $A$ is invertible, this equation has a unique solution and then your equation takes the form
$$ (\mathbf{x} - \mathbf{v})^T A (\mathbf{x} - \mathbf{v}) = \mathbf{v}^T A \mathbf{v}. $$
In order to determine the surface, we must determine the sign of $\mathbf{v}^T A \mathbf{v}$. If the expression is greater than zero, we'll get a one-sheeted hyperboloid while if the expression is less than zero, we'll get a two-sheeted hyperboloid (it won't be zero but if it could be, we'd get a double cone). In your case, we have
$$ \mathbf{v}^T A \mathbf{v} = \frac{11500}{2827} > 0$$
so this is a one-sheeted hyperboloid.
