To solve the equation
\[ bb - 5 - 2b + 5 = 50b^2 - 25, \]
let's rewrite it clearly. It appears you meant \( b^2 - 2b - 5 = 50b^2 - 25 \). If that's correct, we can solve the equation as follows:
- Start with the equation:
\[ b^2 - 2b - 5 = 50b^2 - 25. \]
- Move all terms to one side of the equation by subtracting \( 50b^2 \) and adding \( 25 \):
\[ b^2 - 50b^2 - 2b - 5 + 25 = 0. \]
- Combine like terms:
\[ -49b^2 - 2b + 20 = 0. \]
- Multiply the entire equation by -1 to simplify:
\[ 49b^2 + 2b - 20 = 0. \]
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Now, we can use the quadratic formula \( b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \) where \( A = 49, B = 2, C = -20 \).
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Calculate the discriminant \( B^2 - 4AC \):
\[ B^2 - 4AC = 2^2 - 4 \cdot 49 \cdot (-20). \]
\[ = 4 + 3920 = 3924. \]
- Now, calculate \( \sqrt{3924} \):
Since \( 3924 = 4 \times 981 \) and \( \sqrt{4} = 2 \), we have:
\[ \sqrt{3924} = 2\sqrt{981}. \]
Since \( 981 = 3^2 \times 109 \), rewrite it:
\[ \sqrt{981} = 3\sqrt{109}. \]
So
\[ \sqrt{3924} = 2 \times 3\sqrt{109} = 6\sqrt{109}. \]
- Substitute back into the quadratic formula:
\[ b = \frac{-2 \pm 6\sqrt{109}}{2 \times 49} = \frac{-2 \pm 6\sqrt{109}}{98} = \frac{-1 \pm 3\sqrt{109}}{49}. \]
- Thus, the two solutions for \( b \) are:
\[ b = \frac{-1 + 3\sqrt{109}}{49} \quad \text{and} \quad b = \frac{-1 - 3\sqrt{109}}{49}. \]
These are the solutions to the given equation.