Answer
$$\frac{{{{\left( {x + 1} \right)}^2}}}{{25}} + \frac{{{{\left( {y - 2} \right)}^2}}}{9} = 1$$
Work Step by Step
$$\eqalign{
& 9{x^2} + 25{y^2} + 18x - 100y - 116 = 0 \cr
& {\text{Group terms }} \cr
& 9{x^2} + 18x + 25{y^2} - 100y = 116 \cr
& 9\left( {{x^2} + 2x} \right) + 25\left( {{y^2} - 4y} \right) = 116 \cr
& {\text{Complete the square and factor}} \cr
& 9\left( {{x^2} + 2x + 1} \right) + 25\left( {{y^2} - 4y + 4} \right) = 116 + 9\left( 1 \right) + 25\left( 4 \right) \cr
& 9{\left( {x + 1} \right)^2} + 25{\left( {y - 2} \right)^2} = 225 \cr
& \frac{{9{{\left( {x + 1} \right)}^2}}}{{225}} + \frac{{25{{\left( {y - 2} \right)}^2}}}{{225}} = 1 \cr
& \frac{{{{\left( {x + 1} \right)}^2}}}{{25}} + \frac{{{{\left( {y - 2} \right)}^2}}}{9} = 1 \cr
& {\text{The equation has the form}} \cr
& \frac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} + \frac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1,{\text{ }}a > b \cr
& \underbrace {\frac{{{{\left( {x + 1} \right)}^2}}}{{{{\left( 5 \right)}^2}}} + \frac{{{{\left( {y - 2} \right)}^2}}}{{{{\left( 3 \right)}^2}}} = 1}_ \Downarrow \cr
& h = - 1,k = 2,a = 5,b = 3 \cr
& c = \sqrt {25 - 9} = 4 \cr
& {\text{With}} \cr
& {\text{Vertex }}\left( {h - a,k} \right){\text{ and }}\left( {h + a,k} \right) \cr
& {\text{Vertex }}\left( { - 6,2} \right){\text{ and }}\left( {4,2} \right) \cr
& {\text{Center}}\left( {h,k} \right) \to {\text{Center}}\left( { - 1,2} \right) \cr
& {\text{Foci }}\left( {h - c,k} \right){\text{ and }}\left( {h + c,k} \right) \cr
& {\text{Foci }}\left( { - 5,2} \right){\text{ and }}\left( { - 3,2} \right) \cr
& {\text{Graph}} \cr} $$
