To solve the equation
\[ \frac{b}{b-5} - \frac{2}{b+5} = \frac{50}{b^2-25}, \]
we first notice that \(b^2 - 25\) can be factored:
\[ b^2 - 25 = (b-5)(b+5). \]
Thus, the equation can be rewritten as:
\[ \frac{b}{b-5} - \frac{2}{b+5} = \frac{50}{(b-5)(b+5)}. \]
Next, we find a common denominator for the left side of the equation. The common denominator for \(\frac{b}{b-5}\) and \(\frac{2}{b+5}\) is \((b-5)(b+5)\):
\[ \frac{b(b+5) - 2(b-5)}{(b-5)(b+5)} = \frac{50}{(b-5)(b+5)}. \]
Now we simplify the numerator on the left side:
\[ b(b+5) - 2(b-5) = b^2 + 5b - 2b + 10 = b^2 + 3b + 10. \]
So we have:
\[ \frac{b^2 + 3b + 10}{(b-5)(b+5)} = \frac{50}{(b-5)(b+5)}. \]
Since the denominators are the same, we can set the numerators equal to each other:
\[ b^2 + 3b + 10 = 50. \]
Subtracting 50 from both sides gives:
\[ b^2 + 3b - 40 = 0. \]
Next, we will factor the quadratic equation:
To factor \(b^2 + 3b - 40\), we look for two numbers that multiply to \(-40\) and add to \(3\). The numbers \(8\) and \(-5\) work:
\[ (b + 8)(b - 5) = 0. \]
Setting each factor equal to zero gives us:
\[ b + 8 = 0 \quad \Rightarrow \quad b = -8, \] \[ b - 5 = 0 \quad \Rightarrow \quad b = 5. \]
However, we need to check if these solutions are valid in the original equation, particularly with respect to the denominators:
- For \(b = 5\), the original denominators contain \(b - 5\), which would make the term undefined.
- For \(b = -8\), no denominators are zero.
Thus, the valid solution is:
\[ \boxed{-8}. \]