b(2b-4)=(2b+1)(b+5)

Solution for b(2b-4)=(2b+1)(b+5) equation:

b(2b-4)=(2b+1)(b+5)

We move all terms to the left:

b(2b-4)-((2b+1)(b+5))=0

We multiply parentheses

2b^2-4b-((2b+1)(b+5))=0

We multiply parentheses ..

2b^2-((+2b^2+10b+b+5))-4b=0


We calculate terms in parentheses: -((+2b^2+10b+b+5)), so:

(+2b^2+10b+b+5)

We get rid of parentheses

2b^2+10b+b+5

We add all the numbers together, and all the variables

2b^2+11b+5

Back to the equation:

-(2b^2+11b+5)


We add all the numbers together, and all the variables

2b^2-4b-(2b^2+11b+5)=0

We get rid of parentheses

2b^2-2b^2-4b-11b-5=0

We add all the numbers together, and all the variables

-15b-5=0

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

-15b=5

b=5/-15

b=-1/3