To solve the equation \( bb - 5 - 2b + 5 = 50b^2 - 25 \), we first rewrite it for clarity. Noting that \( bb \) can be interpreted as \( b^2 \), we can rewrite the equation as:
\[ b^2 - 2b = 50b^2 - 25 \]
Next, let's move all terms to one side of the equation to set it equal to zero:
\[ b^2 - 2b - 50b^2 + 25 = 0 \]
Now, combine like terms:
\[ b^2 - 50b^2 - 2b + 25 = 0 \] \[ -49b^2 - 2b + 25 = 0 \]
To simplify solving, we can multiply the entire equation by \(-1\):
\[ 49b^2 + 2b - 25 = 0 \]
Next, we can use the quadratic formula to solve for \( b \). The quadratic formula is given by:
\[ b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \]
In our case, \( A = 49 \), \( B = 2 \), and \( C = -25 \). We need to compute the discriminant \( B^2 - 4AC \):
\[ B^2 - 4AC = 2^2 - 4(49)(-25) \] \[ = 4 + 4900 \] \[ = 4904 \]
Now substituting back into the quadratic formula:
\[ b = \frac{-2 \pm \sqrt{4904}}{2 \cdot 49} \]
Let's simplify \( \sqrt{4904} \): Finding the prime factorization of \( 4904 \):
- \( 4904 = 4 \times 1226 = 4 \times 2 \times 613 = 2^2 \times 2 \times 613 = 2^3 \times 613 \)
Thus, \( \sqrt{4904} = 2\sqrt{1226} \).
Continuing with the square root approximation, we can calculate:
\[ \sqrt{4904} \approx 70 \]
Now substitute this back into the formula:
\[ b = \frac{-2 \pm 70}{98} \]
Calculating the two potential \( b \) values:
- Using the plus sign:
\[ b = \frac{-2 + 70}{98} = \frac{68}{98} = \frac{34}{49} \]
- Using the minus sign:
\[ b = \frac{-2 - 70}{98} = \frac{-72}{98} = \frac{-36}{49} \]
Thus, the solutions to the equation are:
\[ b = \frac{34}{49} \quad \text{or} \quad b = -\frac{36}{49} \]
In summary, the solutions are:
\[ \boxed{\frac{34}{49} \text{ and } -\frac{36}{49}} \]