Let the five numbers be A,B,C,D,E where A ≦ B ≦ C ≦ D ≦ E.

For a set of 5 numbers (or any odd number of numbers), 
the median is the middle number when they are listed
smallest to largest.

so we have  C = 8, and we now have:

                          A ≦ B ≦ 8 ≦ D ≦ E

Those have to be A and B. So we  can insert 0 in between the B and the 8.

                      A ≦ B < 0 < 8 ≦ D ≦ E 

The two negative ones have to be the least two, which are A and B.

That means B - A = 6 or 

(1)  B = A+6, 

so replacing B by A+6 we have:

                     A < A+6 < 0 < 8 ≦ D ≦ E

 So we add them up, divide by 5 and that must equal 18.6:

 

Multiply through by 5 to clear of fractions.

2A+14+D+E = 18.6×5 = 

Simplify:

2A+14+D+E = 93

(2)  2A+D+E = 79       

and the range is 60, so

E-A = 60, or

(3)  E-60 = A

Use (3) to substitute E-60 for A in (2)

(2)  2A+D+E = 79
     2(E-60)+D+E = 79
      2E-120+D+E = 79
            3E+D = 199
(4)            D = 199-3E 


We know that D must be 8 or larger, so

               D ≧ 8

Use (4) to substitute 199-3E for D

          199-3E ≧ 8

Solve for E

             -3E ≧ -191

Dividing through by a negative number, -3, reverses the inequality:

               E ≦ 

So one of D and E is a perfect square and the other is one more than
a perfect square.  So we try perfect squares  and 1 more than perfect
squares less than  and larger than 8, for D and E.

perfect squares:          9, 16, 25, 36, 49
perfect squares plus1:   10, 17, 26, 37, 50 

Let's try D = 49 in (4)

               D = 199-3E  

              49 = 199-3E

              3E = 150

               E = 50

and E=50 is one of those "perfect square + 1's, namely 1 more than 49, or D.

So I think we have it.

Substitute D = 49 and E = 50 in (2)

 (2)  E-60 = A
     50-60 = A
       -10 = A  
     
and substituting in (1)

         B = A+6
         B = -10 + 6
         B = -4.

So the numbers are

A = -10,  B = -4,  C = 8,  D = 49,  E = 50.

Edwin