If function is #y=f(x)# then by the definition #x=f^-1(y)#
To find #f^-1(y)#, we can find the input of #f# that corresponds to an output of #y#.
#f(x)=x^2-3x#
let #f(x)=y#
#f(x)=y=x^2-3x#
Solving equation for #x# we add and subtract #9/4# in the right side of equation to make a perfect square
#y=x^2-3x+9/4-9/4#
#y=(x-3/2)^2-9/4#
Add #9/4# from both side
#y+9/4=(x-3/2)^2-9/4+9/4=(x-3/2)^2#
Taking Square root both side
#sqrt(y+9/4)=sqrt((x-3/2)^2)#
#sqrt(y+9/4)=x-3/2#
Add #3/2# on both side
#+--sqrt(y+9/4)+3/2=x-3/2+3/2#
#+-sqrt(y+9/4)+3/2=x#
#+-sqrt((4y+9)/4)+3/2=x#
#+-sqrt((4y+9))/2+3/2=x#
So
#f^-1(y)=+-sqrt((4y+9))/2+3/2#
Since the choice of the variable is arbitrary, we can write this as
#f^-1(x)=+-sqrt((4x+9))/2+3/2=(3+-sqrt(4x+9))/2#
