$\displaystyle x^2-8x+y^2+2y=-14$
okay this is what I did, I separated the x and y and completed the square for both x and y terms so I resulted in this $\displaystyle (x-4)^2-2+(y+1)^2+13$. I took $\displaystyle x^2-8x+14$ and compeleted the square first and I did the same by adding 14 on the y terms as well.
Hello, Eraser147!
$\displaystyle \text{Complete the square: }\:x^2-8x+y^2+2y\:=\:\text{-}14$
We have: .$\displaystyle x^2 - 8x \qquad + y^2 + 2y \qquad \;=\;\text{-}14$
Then:. . . . $\displaystyle x^2 - 8x {\color{red}\:+\:16} + y^2 + 2y {\color{green}\:+\:1} \;=\;\text{-}14 {\color{red}\:+\:16} {\color{green}\:+\:1} $
Therefore: . . . . . .$\displaystyle (x-4)^2 + (y+1)^2 \;=\;3$