T2(5c+2)-2c=3(2c+3)+7

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Solution for T2(5c+2)-2c=3(2c+3)+7 equation: 



2(5T+2)-2T=3(2T+3)+7

We move all terms to the left:

2(5T+2)-2T-(3(2T+3)+7)=0

We add all the numbers together, and all the variables

-2T+2(5T+2)-(3(2T+3)+7)=0

We multiply parentheses

-2T+10T-(3(2T+3)+7)+4=0



We calculate terms in parentheses: -(3(2T+3)+7), so:

3(2T+3)+7

We multiply parentheses

6T+9+7

We add all the numbers together, and all the variables

6T+16

Back to the equation:

-(6T+16)



We add all the numbers together, and all the variables

8T-(6T+16)+4=0

We get rid of parentheses

8T-6T-16+4=0

We add all the numbers together, and all the variables

2T-12=0

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

2T=12

T=12/2

T=6

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