# How do you find the standard form of x^2 + y^2 + 2x - 6y - 6 = 0 and what kind of a conic is it?

${x}^{2} + 2 x + {y}^{2} - 6 y = 6$
${\left(x + 1\right)}^{2} - 1 + {\left(y - 3\right)}^{2} - 9 = 6$
${\left(x + 1\right)}^{2} + {\left(y - 3\right)}^{2} = 16$