Find the standard form of the equation of the hyperbola with the given characteristics. Vertices: $(-1,2),(5,2)$; foci: $(-3,2),(7,2)$

Step 1 ：Given the vertices and foci of the hyperbola, we can find the center, $h$ and $k$, by averaging the x-coordinates and y-coordinates of the vertices respectively. For vertices $(-1,2)$ and $(5,2)$, we find that $h=\frac{-1+5}{2}=2$ and $k=\frac{2+2}{2}=2$.

Step 2 ：The distance between the vertices, $2a$, is the difference between their x-coordinates, so $a=\frac{5-(-1)}{2}=3$.

Step 3 ：The distance between the foci, $2c$, is the difference between their x-coordinates, so $c=\frac{7-(-3)}{2}=5$.

Step 4 ：We can find $b$ using the relationship $c^2=a^2+b^2$. Solving for $b$, we get $b=\sqrt{c^2-a^2}=\sqrt{25-9}=4$.

Step 5 ：Substituting these values into the standard form of the equation of a hyperbola, we get $\frac{(x-2{)}^2}{9}-\frac{(y-2{)}^2}{16}=1$.

Step 6 ：Final Answer: The standard form of the equation of the hyperbola with the given characteristics is $\boxed{{\displaystyle \frac{(x-2{)}^2}{9}-\frac{(y-2{)}^2}{16}=1}}$.