2(t-3)-32(t-5)=14-5(3t+2)

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Solution for 2(t-3)-32(t-5)=14-5(3t+2) equation: 



2(t-3)-32(t-5)=14-5(3t+2)

We move all terms to the left:

2(t-3)-32(t-5)-(14-5(3t+2))=0

We multiply parentheses

2t-32t-(14-5(3t+2))-6+160=0



We calculate terms in parentheses: -(14-5(3t+2)), so:

14-5(3t+2)

determiningTheFunctionDomain
-5(3t+2)+14

We multiply parentheses

-15t-10+14

We add all the numbers together, and all the variables

-15t+4

Back to the equation:

-(-15t+4)



We add all the numbers together, and all the variables

-30t-(-15t+4)+154=0

We get rid of parentheses

-30t+15t-4+154=0

We add all the numbers together, and all the variables

-15t+150=0

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

-15t=-150

t=-150/-15

t=+10

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