Answer
$$\frac{{{{\left( {x + 4} \right)}^2}}}{2} - \frac{{{{\left( {y - 3} \right)}^2}}}{3} = 1$$
Work Step by Step
$$\eqalign{
& 3{x^2} - 2{y^2} + 24x + 12y + 24 = 0 \cr
& {\text{Group terms}} \cr
& 3{x^2} + 24x - 2{y^2} + 12y = - 24 \cr
& \left( {3{x^2} + 24x} \right) - \left( {2{y^2} - 12y} \right) = - 24 \cr
& 3\left( {{x^2} + 8x} \right) - 2\left( {{y^2} - 6y} \right) = - 24 \cr
& {\text{Complete the square and factor}} \cr
& 3\left( {{x^2} + 8x + 16} \right) - 2\left( {{y^2} - 6y + 9} \right) = - 24 + 3\left( {16} \right) - 2\left( 9 \right) \cr
& 3{\left( {x + 4} \right)^2} - 2{\left( {y - 3} \right)^2} = 6 \cr
& {\text{Divide both sides by 6}} \cr
& \frac{{3{{\left( {x + 4} \right)}^2}}}{6} - \frac{{2{{\left( {y - 3} \right)}^2}}}{6} = 1 \cr
& \frac{{{{\left( {x + 4} \right)}^2}}}{2} - \frac{{{{\left( {y - 3} \right)}^2}}}{3} = 1 \cr
& {\text{This equation has the form}} \cr
& \frac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} - \frac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1 \cr
& {\text{Therefore,}} \cr
& \frac{{{{\left( {x + 4} \right)}^2}}}{2} - \frac{{{{\left( {y - 3} \right)}^2}}}{3} = 1 \Rightarrow \underbrace {\frac{{{{\left( {x + 4} \right)}^2}}}{{{{\left( {\sqrt 2 } \right)}^2}}} - \frac{{{{\left( {y - 3} \right)}^2}}}{{{{\left( {\sqrt 3 } \right)}^2}}} = 1}_{\frac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} - \frac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1} \cr
& h = - 4,{\text{ }}k = 3,{\text{ }}a = \sqrt 2 ,{\text{ }}b = \sqrt 3 \cr
& c = \sqrt {{a^2} + {b^2}} = \sqrt {2 + 3} \Rightarrow c = \sqrt 5 \cr
& {\text{Characteristics of the hyperbola }} \cr
& {\text{Orientation}}:{\text{ Horizontal transverse axis}} \cr
& {\text{Center}}\left( {h,k} \right) \to {\text{Center}}\left( { - 4,3} \right) \cr
& {\text{Vertices}}\left( {h - a,k} \right){\text{ and }}\left( {h + a,k} \right) \cr
& {\text{Vertices}}\left( { - 4 + \sqrt 3 } \right){\text{ and }}\left( { - 4 - \sqrt 3 } \right){\text{ }} \cr
& {\text{Foci}}\left( {h - c,k} \right){\text{ and }}\left( {h + c,k} \right) \cr
& {\text{Foci}}\left( { - 4 + \sqrt 5 ,3} \right){\text{ and }}\left( { - 4 - \sqrt 5 ,3} \right) \cr
& \underbrace {{\text{Asymptotes}}:{\text{ }}y = \pm \frac{b}{a}\left( {x - h} \right) + k}_ \Downarrow \cr
& {\text{Asymptotes}}:{\text{ }}y = \pm \frac{{\sqrt 3 }}{{\sqrt 2 }}\left( {x + 4} \right) + 3 \cr
& {\text{Graph}} \cr} $$
