To solve the equation \( bb - 5 - 2b + 5 = 50b^2 - 25 \), we first rewrite \( bb \) as \( b^2 \). So, the equation becomes:
\[ b^2 - 5 - 2b + 5 = 50b^2 - 25 \]
Next, we can simplify the left-hand side:
\[ b^2 - 2b = 50b^2 - 25 \]
Now, we bring all terms to one side of the equation:
\[ b^2 - 2b - 50b^2 + 25 = 0 \]
Combining like terms, we have:
\[ -49b^2 - 2b + 25 = 0 \]
To make the coefficients more manageable, we can multiply through by -1:
\[ 49b^2 + 2b - 25 = 0 \]
Now, we will use the quadratic formula:
\[ b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \]
In our case, \( A = 49 \), \( B = 2 \), and \( C = -25 \). First, we calculate the discriminant \( B^2 - 4AC \):
\[ B^2 = 2^2 = 4 \] \[ 4AC = 4 \cdot 49 \cdot (-25) = -4900 \]
Now calculate the discriminant:
\[ B^2 - 4AC = 4 + 4900 = 4904 \]
Now we can use the quadratic formula to find \( b \):
\[ b = \frac{-2 \pm \sqrt{4904}}{2 \cdot 49} \]
Calculating \( \sqrt{4904} \):
\[ \sqrt{4904} = \sqrt{4 \cdot 1226} = 2\sqrt{1226} \]
Now substituting back into the quadratic formula:
\[ b = \frac{-2 \pm 2\sqrt{1226}}{98} = \frac{-1 \pm \sqrt{1226}}{49} \]
At this point, we can approximate \( \sqrt{1226} \):
\[ \sqrt{1226} \approx 35 \]
Thus, we have:
\[ b \approx \frac{-1 \pm 35}{49} \]
Calculating the two possible values for \( b \):
- \( b = \frac{-1 + 35}{49} = \frac{34}{49} \)
- \( b = \frac{-1 - 35}{49} = \frac{-36}{49} \)
Therefore, the solutions for \( b \) are:
\[ b \approx \frac{34}{49} \quad \text{and} \quad b \approx \frac{-36}{49} \]
So the final solutions are:
\[ b = \frac{34}{49} \quad \text{and} \quad b = \frac{-36}{49} \]
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