Solve for \( b \):

\[ 4b \times 3b - 2b = 102 \]

4b×3b-2b=102 12b^2 - 2b = 102 12b^2 - 2b - 102 = 0 divide by 2 6b^2 - b - 51 (6b + 17)(b - 3) = 0 6b + 17 = 0 6b = -17 b = -17/6 or 2 5/6 b - 3 = 0 b = 3

Answered by kayt91 | 2024-06-10

4 b ⋅ 3 b − 2 b = 102 12 b 2 − 2 b − 102 = 0 6 b 2 − b − 51 = 0 6 b 2 + 17 b − 18 b − 51 = 0 b ( 6 b + 17 ) − 3 ( 6 b + 17 ) = 0 ( b − 3 ) ( 6 b + 17 ) = 0 b = 3 ∨ b = − 6 17 ​

Answered by konrad509 | 2024-06-10

The equation 4 b × 3 b − 2 b = 102 simplifies to 6 b 2 − b − 51 = 0 . Using the quadratic formula, we find the solutions for b are 3 and − 6 17 ​ .
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Answered by kayt91 | 2024-12-23