TITLE: Why is $4x^2 -3y^2 + 8x +12y-4 =0$ not a valid hyperbola?
QUESTION [1 upvotes]: The equation I want to graph is: $$4x^2 -3y^2 + 8x +12y-4 =0$$
My textbook says that the above equation has no graph. 
Yet, I rearranged it to the following: 
$$\frac{(y-2)^2}{(2/\sqrt3)^2} - \frac{(x+1)^2}{1^2} = 1^2$$
Therefore, I thought it was a hyperbola. I know I've made a mistake in the rearrangement, as the two equations produce different graphs in Desmos. However, both equations produce hyperbolas, which contradicts my textbook. 
Where am I going wrong? Thanks.

REPLY [1 votes]: $4(x+1)^2-3(y-2)^2=-4 \\ \displaystyle \Rightarrow \frac{3(y-2)^2}{4} - (x+1)^2=1$
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