3(t-2)-(3t-4)=2(4t+1)-28

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Solution for 3(t-2)-(3t-4)=2(4t+1)-28 equation: 



3(t-2)-(3t-4)=2(4t+1)-28

We move all terms to the left:

3(t-2)-(3t-4)-(2(4t+1)-28)=0

We multiply parentheses

3t-(3t-4)-(2(4t+1)-28)-6=0

We get rid of parentheses

3t-3t-(2(4t+1)-28)+4-6=0



We calculate terms in parentheses: -(2(4t+1)-28), so:

2(4t+1)-28

We multiply parentheses

8t+2-28

We add all the numbers together, and all the variables

8t-26

Back to the equation:

-(8t-26)



We add all the numbers together, and all the variables

-(8t-26)-2=0

We get rid of parentheses

-8t+26-2=0

We add all the numbers together, and all the variables

-8t+24=0

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

-8t=-24

t=-24/-8

t=+3

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