Convert equations of hyperbolas from general to standard form Algebra II

\dfrac{\left(x - 2\right)^2}4 - \dfrac{\left(y - 2\right)^2} 4 = 1

\dfrac{\left(x - 2\right)^2}{12} - \dfrac{\left(y - 2\right)^2} {12} = 1

\dfrac{\left(x - 2\right)^2}8 - \dfrac{\left(y - 2\right)^2} 8 = 1

\dfrac{\left(x - 2\right)^2}4 + \dfrac{\left(y - 2\right)^2} 4 = 1

\dfrac{\left(x - 3\right)^2}9 - \dfrac{\left(y - 1\right)^2} 9 = 1

\dfrac{\left(x + 2\right)^2}{12} - \dfrac{\left(y - 2\right)^2} {12} = 1

\dfrac{\left(x - 2\right)^2}6 - \dfrac{\left(y - 2\right)^2} 6 = 1

\dfrac{\left(x -3\right)^2}9 + \dfrac{\left(y+ 1\right)^2} 9 = 1

\dfrac{\left(x+1\right)^2}{10}-\dfrac{\left(y-2\right)^2}{10}=1

\dfrac{\left(x - 2\right)^2}{10} - \dfrac{\left(y - 2\right)^2} {10} = 1

\dfrac{\left(x - 3\right)^2}8 - \dfrac{\left(y - 1\right)^2} 8 = 1

\dfrac{\left(x +3\right)^2}8 + \dfrac{\left(y + 2\right)^2} 8 = 1

\dfrac{\left(x - 2\right)^2}{12} - \dfrac{\left(y+1\right)^2} {12} = 1

\dfrac{\left(x + 2\right)^2}{12} - \dfrac{\left(y - 2\right)^2} {12} = 1

\dfrac{\left(x - 2\right)^2}8 - \dfrac{\left(y - 5\right)^2} 8 = 1

\dfrac{\left(x -5\right)^2}8 + \dfrac{\left(y - 2\right)^2} 8 = 1

\dfrac{\left(x +3\right)^2}{25} - \dfrac{\left(y - 2\right)^2} {25} = 1

\dfrac{\left(x - 3\right)^2}{15} - \dfrac{\left(y - 2\right)^2} {15} = 1

\dfrac{\left(x - 2\right)^2}8 - \dfrac{\left(y - 2\right)^2} 8 = 1

\dfrac{\left(x - 2\right)^2}{25} + \dfrac{\left(y - 2\right)^2} {25}= 1

\dfrac{\left(x - 4\right)^2}5 - \dfrac{\left(y + 1\right)^2}5 = 1

\dfrac{\left(x - 2\right)^2}{7} - \dfrac{\left(y - 2\right)^2} {7} = 1

\dfrac{\left(x + 4\right)^2}5 - \dfrac{\left(y - 2\right)^2} 5 = 1

\dfrac{\left(x+ 2\right)^2}7 + \dfrac{\left(y - 2\right)^2} 7 = 1

\dfrac{\left(x - 5\right)^2}7 - \dfrac{\left(y+ 2\right)^2} 7 = 1

\dfrac{\left(x - 5\right)^2}{10} - \dfrac{\left(y - 2\right)^2} {10} = 1

\dfrac{\left(x - 2\right)^2}7 - \dfrac{\left(y - 2\right)^2} 7 = 1

\dfrac{\left(x - 2\right)^2}4 + \dfrac{\left(y - 2\right)^2} 4 = 1