rearrange equation

hi all
please can anyone help with this equation stuck tight with it

need to make b the subject of this equation

7a= b/5 -3c (bc+2).

Any help appreciated please
Eggie

$\displaystyle 7a=\frac{b}{5}-3c(bc+2)=\frac{b}{5}-3bc^2-6c=(\frac{1}{5}-3c^2)b-6c$

Can you do it from there?

Hi RDk

The question i have is this that a student makes a mistake in attempting the equation as follows
7a= b/5 -3(bc+2)

he writes it as this

7a= b/5 - 3 (bc+2)
35a = b-3 (bc-2)
35a = b-3bc + 6
35a - 6 = b(1-3c)

b = 35a - 6 / 1-3c

so somewhere he is wrong i cant figure the equation out so i cant find the mistake that he made. 2 mistakes are made in the attempt of the equation.

so any help appreciated.

eggie