2/5(c+5)+6=7/10(2c+2)

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Solution for 2/5(c+5)+6=7/10(2c+2) equation: 



2/5(c+5)+6=7/10(2c+2)

We move all terms to the left:

2/5(c+5)+6-(7/10(2c+2))=0



Domain of the equation: 5(c+5)!=0

c∈R





Domain of the equation: 10(2c+2))!=0

c∈R



We calculate fractions


(20c2/(5(c+5)*10(2c+2)))+(-35cc/(5(c+5)*10(2c+2)))+6=0



We calculate terms in parentheses: +(20c2/(5(c+5)*10(2c+2))), so:

20c2/(5(c+5)*10(2c+2))

We multiply all the terms by the denominator


20c2

We add all the numbers together, and all the variables

20c^2

Back to the equation:

+(20c^2)





We calculate terms in parentheses: +(-35cc/(5(c+5)*10(2c+2))), so:

-35cc/(5(c+5)*10(2c+2))

We multiply all the terms by the denominator


-35cc

Back to the equation:

+(-35cc)



We get rid of parentheses

20c^2-35cc+6=0

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

20c^2-35cc=-6

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