To solve the equation
\[ \frac{b - 5}{-2b + 5} = \frac{50}{b^2 - 25}, \]
we start by noting that the denominator on the right-hand side can be factored:
\[ b^2 - 25 = (b - 5)(b + 5). \]
We can rewrite the equation as:
\[ \frac{b - 5}{-2b + 5} = \frac{50}{(b - 5)(b + 5)}. \]
Next, we cross-multiply to eliminate the fractions:
\[ (b - 5)(b - 5) = 50(-2b + 5). \]
Now we expand both sides. The left side is:
\[ (b - 5)^2 = b^2 - 10b + 25. \]
The right side simplifies to:
\[ 50(-2b + 5) = -100b + 250. \]
Setting the two sides equal gives us:
\[ b^2 - 10b + 25 = -100b + 250. \]
Now, we will move all terms to one side of the equation:
\[ b^2 - 10b + 25 + 100b - 250 = 0. \]
This simplifies to:
\[ b^2 + 90b - 225 = 0. \]
Now, we can use the quadratic formula \( b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \), where \( A = 1 \), \( B = 90 \), and \( C = -225 \).
Calculating the discriminant:
\[ B^2 - 4AC = 90^2 - 4(1)(-225) = 8100 + 900 = 9000. \]
Now applying the quadratic formula:
\[ b = \frac{-90 \pm \sqrt{9000}}{2}. \]
The square root of \( 9000 \) can be simplified:
\[ \sqrt{9000} = \sqrt{90^2 \times 10} = 90 \sqrt{10}. \]
Thus, we have:
\[ b = \frac{-90 \pm 90 \sqrt{10}}{2} = -45 \pm 45 \sqrt{10}. \]
Therefore, the final solutions for \( b \) are:
\[ b = -45 + 45\sqrt{10} \quad \text{and} \quad b = -45 - 45\sqrt{10}. \]
These are the solutions to the equation.