Quadratic solution of the Migration / Selection
            equilibrium
            
          To solve  qi  = tqi2 - (m + t)(qi) + mqm
qi  = tqi2 - (m + t)(qi) + mqm
    
    
      Recall the quadratic formula: 0
          = [-b 
 (b2 - 4ac)] / 2a
(b2 - 4ac)] / 2a
    
      Then, set terms of quadratic as   a
      = t    b
      = - (m + t)   c
      = mqm 
                                   
    
         = [ (m + t)
= [ (m + t) 
 [( (-1)2(m + t)2
      - 4tmqm]
      ] / 2t         
      quadratic form
[( (-1)2(m + t)2
      - 4tmqm]
      ] / 2t         
      quadratic form
    
            =  [ (m + t) 
 [m2 + t2
      + 2mt - 4mt] ] / 2t         
          qm =
          1; expand ( - (m + t) )2
[m2 + t2
      + 2mt - 4mt] ] / 2t         
          qm =
          1; expand ( - (m + t) )2
     
             [ (m + t)
  [ (m + t) 
 [t2 +
      2mt - 4mt] ] / 2t              
             if 
        m < t , then m2 << t2
[t2 +
      2mt - 4mt] ] / 2t              
             if 
        m < t , then m2 << t2 
     
    
             [
      (m + t)
  [
      (m + t) 
 [t2
      + (t2)(2m/t - 4m/t)] ] / 2t    
             create common t2 term
[t2
      + (t2)(2m/t - 4m/t)] ] / 2t    
             create common t2 term  
    
             [
      (m + t)
  [
      (m + t)  (t)
 (t)  [1 + 2m/t - 4m / t] ] / 2t    
               factor out
[1 + 2m/t - 4m / t] ] / 2t    
               factor out   t2 term as t
t2 term as t 
    
             [ (m + t)
  [ (m + t)  (t)(1
          + m/t - 2m/t) ] /
          2t              
              apply approximation
 (t)(1
          + m/t - 2m/t) ] /
          2t              
              apply approximation
                                                                             
                     (1
(1  x)
 x)    ( 1
  ( 1  (x/2))
 (x/2)) 
                                                                                   
                      because (1 +
            x/2)2  =  1 + x + 
            (x/4)2 
                           
                           
                           
                           
                         
                          
                         (1 + x)    iff x << 1
  (1 + x)    iff x << 1
                                                                                                                                
          that is (- m / t) << 1
         
        For which the negative root, if  t >
          0, is 
    
        
 
 
         (m + t) - (t + m
        - 2m) / 2t              
                 t & m
        terms cancel [ pretty !] 
    
        
 2m / 2t  =  (m / t)   
                           
                    if qm
          = 1 as in standard model, & m < t
 
        2m / 2t  =  (m / t)   
                           
                    if qm
          = 1 as in standard model, & m < t