Answers: 2
![x^2+y^2+4x-6y+12=0\\\\x^2+2x\cdot2+2^2+y^2-2y\cdot3+3^2-2^2-3^2+12=0\\\\(x+2)^2+(y-3)^2-4-9+12=0\\\\(x+2)^2+(y-3)^2-1=0\\\\(y-3)^2=1-(x+2)^2\\\\(y-3)^2=[1-(x+2)][1+(x+2)]\\\\(y-3)^2=(-x-1)(x+3)\\\\y-3=\sqrt{-(x+1)(x+3)}\\\\y=3+\sqrt{-(x+1)(x+3)}](/tpl/images/0194/3955/18ab9.png)
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Given the equation x^2+y^2+4x-6y+12=0 solve for y...
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