Identify and sketch the surface x^2+y^2-4z^2+4x-6y-8z=0

identify and sketch the surface
x^2+y^2-4z^2+4x-6y-8z=0

i get:
-(x+2)^2/3.5-(y-3)^2/3.5-(z+1)^2/3.5=1
(-,-,-,1)
so is that the same as an ellipsoid (+++1)?

x² + y² - 4z² + 4x - 6y - 8z = 0 implies
(x² + 4x + 4) + (y² - 6y + 9) - 4(z² + 2z + 1) = 4 + 9 - 4, or
(x + 2)² + (y - 3)² - 4(z + 1)² = 3²
This resembles the canonic equation
(x + 2)²/6² + (y - 3)²/6² - (z + 1)²/3² = 1
of a hyperboloid of revolution of 1 sheet:
http://en.wikipedia.org/wiki/Hyperboloid…
The radius of the 'throat' is 6, the center is in (-2, 3, -1).

Looks right, but it isn't quite, must be:
(x + 2)²/3² + (y - 3)²/3² - (z + 1)²/(3/2)² = 1
Sorry, I have overlooked that '4', the surface is a hyperboloid of revolution indeed, but the radius of the throat is 3, not 6. Sorry again!

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No, it's a cone. A cone has the general form (z-c)^2 = (x-a)^2 + (y-b)^2 + d. By completing the squares, the equation above can be written in this manner.