Question 753610
{{{x^2-8x-2y^2-12y-4=0}}} --> {{{x^2-8x-2y^2-12y=4}}} --> {{{x^2-8x+16-2y^2-12y-18=4+16-18}}} --> {{{(x^2-8x+16)-2(y^2+6y+9)=4+16-18}}} --> {{{(x-4)^2-2(y+3)^2=2}}}
Now that the squares are completed, we could divide both sides of the equal sign by 2 to get
{{{(x-4)^2-2(y+3)^2=2}}} --> {{{(x-4)^2/2-(y+3)^2/1=1}}}
and we would know that the equation represents a hyperbola with vertices at
({{{4-sqrt(2)}}},{{{-3}}}) and ({{{4+sqrt(2)}}},{{{-3}}})