How do I solve this equation? 5b³+34b²=7b

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5b^3+34b^2=7b is 5b^3+34b^2-7b=0 which is b(5b^2+34b-7)=0.
So b=0 is one solution. 5b^2+34b-7 factorises, so (5b-1)(b+7)=0 and b=1/5 or -7.
The solution is b=-7, 0, 1/5.
by Top Rated User (1.2m points)

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