2(t+2)-(3t-5)=5(2t+3)-28

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Solution for 2(t+2)-(3t-5)=5(2t+3)-28 equation: 



2(t+2)-(3t-5)=5(2t+3)-28

We move all terms to the left:

2(t+2)-(3t-5)-(5(2t+3)-28)=0

We multiply parentheses

2t-(3t-5)-(5(2t+3)-28)+4=0

We get rid of parentheses

2t-3t-(5(2t+3)-28)+5+4=0



We calculate terms in parentheses: -(5(2t+3)-28), so:

5(2t+3)-28

We multiply parentheses

10t+15-28

We add all the numbers together, and all the variables

10t-13

Back to the equation:

-(10t-13)



We add all the numbers together, and all the variables

-1t-(10t-13)+9=0

We get rid of parentheses

-1t-10t+13+9=0

We add all the numbers together, and all the variables

-11t+22=0

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

-11t=-22

t=-22/-11

t=+2

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