# Is $\{(x,y)\in \Bbb R^2: x^2+4y^2+6x+4y+12=0\}$ the empty set?

I am attempting the following problem and can someone verify it ?

We see that $\,x^2+4y^2+6x+4y+12=0 \implies (x+3)^2+4(y+\frac 12)^2=-2$ and hence option (d) is the correct choice. Am I right?

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Is there any particular reason you were unsure? – Clayton Oct 23 '13 at 16:53
I removed the "elementary set theory" tag because the question boils down to "does this equation have any solutions?" (and it seems like the OP knows this.) Please feel free to revert if the notion of "empty set" really was important to the question. – Trevor Wilson Oct 23 '13 at 17:04