SOLUTION: Solving by substitution with u {{{ x^2-3x-sqrt(x^2-3x)=2 }}} {{{ u = sqrt(x^2-3x) }}} {{{ u^2 = sqrt(x^2-3x) = x^2-3x }}} creates quadratic = {{{ u^2-u = 2 }}} {{{ u^2-u-2=0

Solving by substitution with u




creates quadratic = 


u = 2,-1
going back and using those two outputs in 
 and 
 and 
Finally, I get  and 

But -2 can't be in a radical and would 4 turn into 2,-2 or just 2? 
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u = 2,-1 <=== This is okay!
"going back and using those two outputs in      and  <=== Here's where I guess 
                                                                                                                you got confused, and substituted 2 and - 1 for x in 

But, u = 2, as you mentioned above, NOT x = 2. And, because you'd substituted u for  earlier, at this juncture, you
need to BACK-SUBSTITUTE the value of u to get:
                       
                    ---- Squaring both sides
                    
     
(x - 4)(x + 1) = 0
  x - 4 = 0          OR        x + 1 = 0
       x = 4           OR              x = - 1 

Now, you have 2 values for x that you can CHECK to ensure that they're VALID and NOT EXTRANEOUS. 

Also, u = - 1, as you mentioned above. And, because you'd substituted u for  earlier, at this juncture, you need 
to BACK-SUBSTITUTE the value of u to get:  
Seeing that the square root of ANY expression is positive (> 0), it's obvious that u = - 1 is an EXTRANEOUS value. As such,
x = 4, or x = - 1 (see above).