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(2b-3)(3b-8)=0 am i correct with the answer of -6 and 8 for being the solution to this?
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Try substituting in the solutions: (2(-6) -3)(3(-6) - 8) = 0 (-12-3)(-18-8)=0 (-15)(-26)=0 So they are not the solutions.
Instead set each factor equal to 0 and solve: 2b-3 = 0 and 3b-8 = 0
i dotn get what the solutions would be somehow i keep coming up with -6 and 8
2b - 3 = 0 add 3 to both sides 2b = 3 now divide by 2 b = 3/2
I think you were multiplying on the last step instead of dividing by mistake.
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so 3/2 is one solution so what would the second solution be
b = 3/2 , b= 8/3
Try to solve the other equation we had earlier: 3b - 8 = 0
ok
3z^2+12z=0 one solution would be 0 is that the only solution
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yeah
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