IIT JEE , AIEEE Forum - Mathematics , Algebra - find total no. of real solutions of x^2 - 4

LAMPARD

See,
x²-4-x+{x}=0 where {x} is frac. part of x.
Thus,it follows that 0<-x²+4+x<1
So,you get x²-x-4<0           and     x²-x-3>0
              (x-1/2)²<17/4       and     (x-1/2)²>13/4
Thus,      x<(

17+1)/2 x>(1+

13)/2
or and or
x>(1-

17)/2 x<(1-

13)2
Thus,     x<2.56                               x<-1.303
                or                    and             or
              x>-1.56                            x>2.3
Thus x has interval (-1.56,-1.303)U(2.3,2.56)
But i am not able to find the 2 solutions to it...

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