Let \( d \) be the number of days it takes for the original 5 workers to complete the job. The work done by the 5 workers in \( d \) days can be expressed in terms of a single worker's rate of work.
Let the work done by one worker in one day be \( w \). Therefore, the total amount of work \( W \) needed to complete the job is:
\[ W = 5 \times d \times w \]
If one additional worker is hired, making a total of 6 workers, the amount of time they take to complete the job is \( d - 8 \) days. The work done by the 6 workers in this time is:
\[ W = 6 \times (d - 8) \times w \]
Since both expressions equal the same amount of work, we can set them equal to each other:
\[ 5d \cdot w = 6(d - 8) \cdot w \]
Assuming \( w \neq 0 \) (since the workers are doing some work), we can divide both sides by \( w \):
\[ 5d = 6(d - 8) \]
Expanding the right side:
\[ 5d = 6d - 48 \]
Now, isolate \( d \):
\[ 5d - 6d = -48 \]
\[ -d = -48 \]
\[ d = 48 \]
Now that we know it takes 5 workers 48 days to complete the job, we can calculate the total amount of work:
\[ W = 5 \times 48 \times w = 240w \]
Next, we find out how many additional workers are needed to finish the job 28 days earlier than the original timeline, meaning the job must be completed in \( 48 - 28 = 20 \) days.
Let \( x \) be the total number of workers needed to complete the job in 20 days. The equation for the work done by \( x \) workers in \( 20 \) days is:
\[ W = x \times 20 \times w \]
Setting this equal to the total work \( W \):
\[ 240w = x \times 20 \times w \]
Again, assuming \( w \neq 0 \), we can divide both sides by \( w \):
\[ 240 = 20x \]
Now, solve for \( x \):
\[ x = \frac{240}{20} = 12 \]
Since we initially had 5 workers, the number of additional workers needed is:
\[ 12 - 5 = 7 \]
Thus, the number of additional workers that should be hired is:
\[ \boxed{7} \]