Answer
$$\frac{{{{\left( {x - 3} \right)}^2}}}{4} - \frac{{{{\left( {y + 2} \right)}^2}}}{9} = 1$$
Work Step by Step
$$\eqalign{
& {\text{vertices at }}\left( {5, - 2} \right),\left( {1, - 2} \right);{\text{ asymptotes }}y = \pm \frac{3}{2}\left( {x - 3} \right) - 2 \cr
& {\text{The }}y{\text{ - coordinate in the vertices are the same, then the hyperbola}} \cr
& {\text{has the equation }}\frac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} - \frac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1 \cr
& {\text{With vertices at }}\left( {h \pm a,k} \right){\text{ and asymptotes }}y = \pm \frac{b}{a}\left( {x - h} \right) + k \cr
& {\text{,then}} \cr
& {\text{vertices }}\left( {5, - 2} \right),\left( {1, - 2} \right) \to \,\,\,k = - 2 \cr
& {\text{asymptotes }}y = \pm \frac{3}{2}\left( {x - 3} \right) - 2 \to h = 3 \cr
& h + a = 5 \cr
& 3 + a = 5 \cr
& a = 2 \cr
& \cr
& \frac{b}{a} = \frac{3}{2} \cr
& \frac{b}{2} = \frac{3}{2} \cr
& b = 3 \cr
& \cr
& {\text{The equation of the hyperbola is}} \cr
& \frac{{{{\left( {x - 3} \right)}^2}}}{{{2^2}}} - \frac{{{{\left( {y + 2} \right)}^2}}}{{{3^2}}} = 1 \cr
& \frac{{{{\left( {x - 3} \right)}^2}}}{4} - \frac{{{{\left( {y + 2} \right)}^2}}}{9} = 1 \cr} $$
