Question

 given the equation x^2+y^2+4x-6y+12=0 solve for y

Answer 1

$x^2+y^2+4x-6y+12=0\\\\x^2+2x\cdot2+2^2+y^2-2y\cdot3+3^2-2^2-3^2+12=0\\\\(x+2)^2+(y-3)^2-4-9+12=0\\\\(x+2)^2+(y-3)^2-1=0\\\\(y-3)^2=1-(x+2)^2\\\\(y-3)^2=[1-(x+2)][1+(x+2)]\\\\(y-3)^2=(-x-1)(x+3)\\\\y-3=\sqrt{-(x+1)(x+3)}\\\\y=3+\sqrt{-(x+1)(x+3)}$

Answer 2

$x^2+y^2+4x-6y+12=0 \\ y^2-6y+9=-x^2-4x-3\\ (y-3)^2=-x^2-4x-3\\ y-3=\sqrt{-x^2-4x-3} \vee y-3=-\sqrt{-x^2-4x-3}\\ y=3+\sqrt{-x^2-4x-3} \vee y=3-\sqrt{-x^2-4x-3}$