Question 89064


  solve:

5(x-2)^2=3        consider 5(x-2)^2 = 3
                           5(x^2-4x+4)-3 = 0
                           5x^2-20x+20-3 = 0
                           5x^2-20x+17  = 0

                        use the standard formula  x = -b+-sq rtb^2-4ac/2a

                       where a = 5 b = -20  c = 17

                      x = -(-20)+-sq rt(-20)^2-4(5)(17)/2.5
                        = 40+-sq rt(400-340)/10
                        = 40+-sq rt60/10
                        = 40+-2sq rt15/10
                        = 2(20+-sq rt15)/10
                        = (20+-sq rt15)/5  is the required  solution