To solve the equation \( bb - 5 - 2b + 5 = 50b^2 - 25 \), we first rewrite \( bb \) as \( b^2 \). Thus, the equation becomes:
\[ b^2 - 5 - 2b + 5 = 50b^2 - 25 \]
Now, simplify the left side:
\[ b^2 - 2b + 0 = 50b^2 - 25 \]
This simplifies to:
\[ b^2 - 2b = 50b^2 - 25 \]
Next, we move all terms to one side of the equation. Subtract \( 50b^2 - 25 \) from both sides:
\[ b^2 - 2b - 50b^2 + 25 = 0 \]
Combining like terms gives:
\[ -b^2 - 2b + 25 = 0 \]
To make it standard, multiply the entire equation by -1:
\[ b^2 + 2b - 25 = 0 \]
Now we can solve this quadratic equation using the quadratic formula:
\[ b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \]
Where \( A = 1 \), \( B = 2 \), and \( C = -25 \). Calculating the discriminant:
\[ B^2 - 4AC = 2^2 - 4 \cdot 1 \cdot (-25) = 4 + 100 = 104 \]
Now substitute back into the formula:
\[ b = \frac{-2 \pm \sqrt{104}}{2 \cdot 1} \]
\(\sqrt{104}\) can be simplified to \( 2\sqrt{26} \), thus we have:
\[ b = \frac{-2 \pm 2\sqrt{26}}{2} \] \[ b = -1 \pm \sqrt{26} \]
Hence, the two possible solutions for \( b \) are:
\[ b = -1 + \sqrt{26} \quad \text{and} \quad b = -1 - \sqrt{26} \]
Therefore, the final solutions are:
\[ \boxed{-1 + \sqrt{26}} \quad \text{and} \quad \boxed{-1 - \sqrt{26}} \]