SOLUTION: Find the real solutions of each equation. Let rt = root 1. x^2 - 3x - rt{x^2 - 3x} = 2 2. 3x^(3/4) + 5x^(2/3) - 2 = 0

    x^2 - 3x -  = 2.    (1)


It is a standard equation to solve using "change of a variable".


So, we introduce new variable  y = .


Then equation (1) takes the form

    y^2 - y = 2,

or

    y^2 - y - 2 = 0.


We look for non-negative solutions of this equation.


Factor left side

    (y+1)*(y-2) = 0


and get two roots y= -1  and  y= 2.


For what follows, we consider only positive value y= 2.

The root y= 2  leads  to equation 

    x^2 - 3x = 4,

    x^2 - 3x - 4 = 0,

    (x-4)*(x+1) =  0   with the solutions  x= 4  and x= -1.


Of these two solutions, both work. 


ANSWER.  There are two real solutions for x: they are  x= -1  and  x= 4.