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 \\ y^2-6y+9=-x^2-4x-3\\ (y-3)^2=-x^2-4x-3\\ y-3=√(-x^2-4x-3) \vee y-3=-√(-x^2-4x-3)\\ y=3+√(-x^2-4x-3) \vee y=3-√(-x^2-4x-3)`

Answer 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=√(-(x+1)(x+3))\\\\y=3+√(-(x+1)(x+3))`