Determine the equation of the second order curve

Question

I am trying to transform the equation of the second order curve to its canonical form and determine the type of the curve and plot its graph $2x^2+y^2+4x-6y+11=0$ So I tried to use factoring polynomials And got this form $2(x+1)^2+(y-3)^2 = 0$ and actually I know $9 $ equations from the conic section but this doesn't look like one of them ! So I can't determine the type or graph it where is the mistake!

Answer 1

Now that you have updated the constant term, you have the correct form $$2x^2+y^2+4x-6y+11=0\\2(x+1)^2+(y-3)^2=0$$ When a sum of squares is zero, each square must be zero, so we can say $$2(x+1)^2=0\\(y-3)^2=0\\x=-1\\y=3$$

so the only solution is the point $(-1,3)$