SOLUTION: Solve the equation [x² + 2x + 5] - [x + 1] = 6, where the symbol [ ] denotes the greatest integer function (floor function).

For clarity, I will write this equation in this form

    floor(x^2 + 2x + 5) - floor(x+1) = 6.    (1)


I will solve it step by step.



(1)  x^2 + 2x + 5 = (x+1)^2 + 4.


     Therefore,  floor(x^2 + 2x + 5) = floor((x+1)^2 + 4).

     Next, it is clear that  floor((x+1)^2+4) = floor((x+1)^2) + 4.


     Therefore, equation (1) is equivalent to 

         floor((x+1)^2) + 4 - floor(x+1) = 6,

     or

         floor((x+1)^2 - floor(x+1) = 2.



(2)  Now I will introduce new variable  u = x+1  to simplify writing.

     So, now an equation to solve is

         floor(u^2) - floor(u) = 2.    (2)



(3)  For what will follow, it is convenient to look at the plot of the functions  

          u --> y = floor(u^2) - floor(u)  and  y = 2.


     These functions are shown in the plot  https://www.desmos.com/calculator/9eg9wmaonb



(4)  In the interval  -1 < u < 0,  floor(u^2) = 0  and  floor(u) = -1,  so  floor(u^2) - floor(u) = 0 - (-1) = 1,

                                    which means that there are no solutions to equation (2) in this interval.



(5)  At point u = -1,  floor(u^2) = 1;  floor(u) = -1,  so  floor(u^2) - floor(u) = 1 - (-1) = 2,

                                    which means that u = -1  is a solution to equation (2).



(5)  In the interval  -2 < u < -1,  floor(u^2)  has values 1 or 2 or 3;   floor(u) = -2,  

         so  floor(u^2) - floor(u)  has values 3, or 4, or 5, 

                                    which means that there are no solutions to equation (2) in this interval.


                    At this point, it is clear that there is no sense to analyze further 
                    for negative 'u' lesser than  -2, because there are no solutions to equation (2) there.



(6)  In the interval  0 < u < 1,  floor(u^2) has value 0;  floor(u)  has value 0, so  floor(u^2) - floor(u) = 0 - 0 = 0,

                                    which means that there are no solutions to equation (2) in this interval.



(7)  At point u = 1,  floor(u^2) = 1;  floor(u) = 1,  so  floor(u^2) - floor(u) = 1 - 1 = 0,

                                    which means that u = 1  is not a solution to equation (2).



(8)  In the interval  1 < u < ,  floor(u^2)  has value1 1;   floor(u) = 1,  

         so  floor(u^2) - floor(u) = 1 - 1 = 0, 

                                    which means that there are no solutions to equation (2) in this interval.



(9)  At point u = ,  floor(u^2) = 2;  floor(u) = 1,  so  floor(u^2) - floor(u) = 2 - 1 = 1,

                                    which means that u =   is not a solution to equation (2).



(10)  In the interval   < u < ,  floor(u^2)  has value 2;   floor(u) = 1,  

         so  floor(u^2) - floor(u) = 2 - 1 = 1, 

                                    which means that there are no solutions to equation (2) in this interval.



(11)  At point u = ,  floor(u^2) = 3;  floor(u) = 1,  so  floor(u^2) - floor(u) = 3 - 1 = 2,

                                    which means that u =   is a solution to equation (2).



(12)  In the interval   <= u < 2,  floor(u^2)  has value 3;   floor(u) = 1,  

         so  floor(u^2) - floor(u) = 3 - 1 = 2, 

                                    which means that all this interval is the solution to equation (2).



(13)  At point u = 2,  floor(u^2) = 4;  floor(u) = 2,  so  floor(u^2) - floor(u) = 4 - 2 = 2,

                                    which means that u = 2  is a solution to equation (2).



(14)  In the interval  2 < u < ,  floor(u^2)  has value 4;   floor(u) = 2,  

         so  floor(u^2) - floor(u) = 4 - 2 = 2, 

                                    which means that all this interval is the solution to equation (2).



(15)  At point u = ,  floor(u^2) = 5;  floor(u) = 2,  so  floor(u^2) - floor(u) = 5 - 2 = 3,

                                    which means that u =   is not a solution to equation (2).


                    At this point, it is clear that there is no sense to analyze further 
                    for positive 'u' greater than  , because there are no solutions to equation (2) there.



At this point, we can summarize the solution set for variable u:  it is  the union of sets of real numbers  

    < -1 >  U  [  <= u <  ).


Now we can return to variable x and to describe the solution set for the given equation (1) in terms of x.  It is

    < -2 >  U  [  <= x <  ).    <<<---===  ANSWER