(b+4)(b-3)=(b-1)(b+5)+2(b-2)

Solution for (b+4)(b-3)=(b-1)(b+5)+2(b-2) equation:

(b+4)(b-3)=(b-1)(b+5)+2(b-2)

We move all terms to the left:

(b+4)(b-3)-((b-1)(b+5)+2(b-2))=0

We multiply parentheses ..

(+b^2-3b+4b-12)-((b-1)(b+5)+2(b-2))=0


We calculate terms in parentheses: -((b-1)(b+5)+2(b-2)), so:

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

We multiply parentheses

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

We multiply parentheses ..

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

We get rid of parentheses

b^2+5b-1b+2b-5-4

We add all the numbers together, and all the variables

b^2+6b-9

Back to the equation:

-(b^2+6b-9)


We get rid of parentheses

b^2-b^2-3b+4b-6b-12+9=0

We add all the numbers together, and all the variables

-5b-3=0

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

-5b=3

b=3/-5

b=-3/5