Finding inverses (again)

Is #7 this: \(y=\frac{x^2-3}{2}\)

7.   

 y  = √[2x - 3]     square both sides

y^2  = 2x  - 3

y^2 + 3   = 2x     

[ y^2 + 3 ] / 2  =x

[x^2  + 3] / 2    = y

8. 

y = √[4x] - 7

y + 7  =√[4x]   square both sides

(y + 7)^2  = 4x

( y + 7)^2  / 4   = x

(x + 7)^2 / 4   = y