B(k-6)+58=2(4k+5)

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Solution for B(k-6)+58=2(4k+5) equation: 



(B-6)+58=2(4B+5)

We move all terms to the left:

(B-6)+58-(2(4B+5))=0

We get rid of parentheses

B-(2(4B+5))-6+58=0



We calculate terms in parentheses: -(2(4B+5)), so:

2(4B+5)

We multiply parentheses

8B+10

Back to the equation:

-(8B+10)



We add all the numbers together, and all the variables

B-(8B+10)+52=0

We get rid of parentheses

B-8B-10+52=0

We add all the numbers together, and all the variables

-7B+42=0

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

-7B=-42

B=-42/-7

B=+6

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