Q29-(2c+3)=2(c+3)+c

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Solution for Q29-(2c+3)=2(c+3)+c equation: 



29-(2Q+3)=2(Q+3)+Q

We move all terms to the left:

29-(2Q+3)-(2(Q+3)+Q)=0

We get rid of parentheses

-2Q-(2(Q+3)+Q)-3+29=0



We calculate terms in parentheses: -(2(Q+3)+Q), so:

2(Q+3)+Q

We add all the numbers together, and all the variables

Q+2(Q+3)

We multiply parentheses

Q+2Q+6

We add all the numbers together, and all the variables

3Q+6

Back to the equation:

-(3Q+6)



We add all the numbers together, and all the variables

-2Q-(3Q+6)+26=0

We get rid of parentheses

-2Q-3Q-6+26=0

We add all the numbers together, and all the variables

-5Q+20=0

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

-5Q=-20

Q=-20/-5

Q=+4

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