2(t-3)=3(t-2)=28

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



2(t-3)=3(t-2)=28

We move all terms to the left:

2(t-3)-(3(t-2))=0

We multiply parentheses

2t-(3(t-2))-6=0



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

3(t-2)

We multiply parentheses

3t-6

Back to the equation:

-(3t-6)



We get rid of parentheses

2t-3t+6-6=0

We add all the numbers together, and all the variables

-t=0

t=0/-1

t=0

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