Answer
$$\eqalign{
& \,{\text{Vertex }}\left( {h,k} \right) \to {\text{Vertex }}\left( {2,5} \right) \cr
& \,\,\,\,\,\,\,{\text{Horizontal axis of symmetry}} \cr
& \,\,\,\,\,\,\,{\text{Domain: }}\left[ {h,\infty } \right):\left[ {\frac{7}{4},\infty } \right) \cr
& \,\,\,\,\,\,\,{\text{Range:}}\left( { - \infty ,\infty } \right) \cr} $$
Work Step by Step
$$\eqalign{
& x = 5{y^2} - 5y + 3 \cr
& x - 3 = 5{y^2} - 5y \cr
& x - 3 = 5\left( {{y^2} - y} \right) \cr
& x - 3 + 5\left( {\frac{1}{4}} \right) = 5\left( {{y^2} - y + \frac{1}{4}} \right) \cr
& x - \frac{7}{4} = 5{\left( {y - \frac{1}{2}} \right)^2} \cr
& {\left( {y - \frac{1}{2}} \right)^2} = \frac{1}{5}\left( {x - \frac{7}{4}} \right) \cr
& {\text{The equation is written in the form }}{\left( {y - k} \right)^2} = 4p\left( {x - h} \right) \cr
& {\left( {y - \frac{1}{2}} \right)^2} = \frac{1}{5}\left( {x - \frac{7}{4}} \right) \to k = \frac{1}{2},\,\,\,h = \frac{7}{4} \cr
& \cr
& {\text{With: }} \cr
& \,\,\,\,\,\,\,\,{\text{Vertex }}\left( {h,k} \right) \to {\text{Vertex }}\left( {2,5} \right) \cr
& \,\,\,\,\,\,\,{\text{Horizontal axis of symmetry}} \cr
& \,\,\,\,\,\,\,{\text{Domain: }}\left[ {h,\infty } \right):\left[ {\frac{7}{4},\infty } \right) \cr
& \,\,\,\,\,\,\,{\text{Range:}}\left( { - \infty ,\infty } \right) \cr} $$
