740+32(n-10)=1000+25(n-12)

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Solution for 740+32(n-10)=1000+25(n-12) equation: 



740+32(n-10)=1000+25(n-12)

We move all terms to the left:

740+32(n-10)-(1000+25(n-12))=0

We multiply parentheses

32n-(1000+25(n-12))-320+740=0



We calculate terms in parentheses: -(1000+25(n-12)), so:

1000+25(n-12)

determiningTheFunctionDomain
25(n-12)+1000

We multiply parentheses

25n-300+1000

We add all the numbers together, and all the variables

25n+700

Back to the equation:

-(25n+700)



We add all the numbers together, and all the variables

32n-(25n+700)+420=0

We get rid of parentheses

32n-25n-700+420=0

We add all the numbers together, and all the variables

7n-280=0

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

7n=280

n=280/7

n=40

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