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