Respuesta :

konrad509
[tex] 4b\cdot3b-2b=102\\ 12b^2-2b-102=0\\ 6b^2-b-51=0\\ 6b^2+17b-18b-51=0\\ b(6b+17)-3(6b+17)=0\\ (b-3)(6b+17)=0\\\ b=3 \vee b=-\frac{17}{6}[/tex]
kayt91
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

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