(b+1)(b+4)=b(b+3)-8

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Solution for (b+1)(b+4)=b(b+3)-8 equation: 



(b+1)(b+4)=b(b+3)-8

We move all terms to the left:

(b+1)(b+4)-(b(b+3)-8)=0

We multiply parentheses ..

(+b^2+4b+b+4)-(b(b+3)-8)=0



We calculate terms in parentheses: -(b(b+3)-8), so:

b(b+3)-8

We multiply parentheses

b^2+3b-8

Back to the equation:

-(b^2+3b-8)



We get rid of parentheses

b^2-b^2+4b+b-3b+4+8=0

We add all the numbers together, and all the variables

2b+12=0

We move all terms containing b to the left, all other terms to the right

2b=-12

b=-12/2

b=-6

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