(frac{1}{(b-5)^{2}}-frac{1}{(b-5)}-6=0)

To solve this equation, we can first make a substitution to simplify it. Let's substitute x = (b-5), so the equation becomes:

1/x^2 - 1/x - 6 = 0

Now, we can rewrite this equation as a quadratic equation by multiplying through by x^2:

1 - x - 6x^2 = 0

Rearranging the terms, we get:

6x^2 + x - 1 = 0

Now, we can solve this quadratic equation using the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / 2a

In this case, a = 6, b = 1, and c = -1. Plugging these values into the formula, we get:

x = (-1 ± √(1^2 - 4(6)(-1))) / 2(6) x = (-1 ± √(1 + 24)) / 12 x = (-1 ± √25) / 12 x = (-1 ± 5) / 12

So, x = 4/3 or x = -1/2.

Now, we can substitute back x = (b-5) into these solutions to find the values of b:

For x = 4/3: b - 5 = 4/3 b = 4/3 + 5 b = 19/3

For x = -1/2: b - 5 = -1/2 b = -1/2 + 5 b = 9/2

Therefore, the solutions to the original equation are b = 19/3 and b = 9/2.