3(2-6)/2=-4(2c+3)/7

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Solution for 3(2-6)/2=-4(2c+3)/7 equation: 



3(2-6)/2=-4(2c+3)/7

We move all terms to the left:

3(2-6)/2-(-4(2c+3)/7)=0

We add all the numbers together, and all the variables

-(-4(2c+3)/7)+3(-4)/2=0

We calculate fractions


(-(-4(2c+3)*2)/()+()/()=0



We calculate terms in parentheses: +(-(-4(2c+3)*2)/()+()/(), so:

-(-4(2c+3)*2)/()+()/(

We add all the numbers together, and all the variables

-(-4(2c+3)*2)/()+1

We multiply all the terms by the denominator


-(-4(2c+3)*2)+1*()



We calculate terms in parentheses: -(-4(2c+3)*2), so:

-4(2c+3)*2

We multiply parentheses

-16c-24

Back to the equation:

-(-16c-24)



We add all the numbers together, and all the variables

-(-16c-24)

We get rid of parentheses

16c+24

Back to the equation:

+(16c+24)



We get rid of parentheses

16c+24=0

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

16c=-24

c=-24/16

c=-1+1/2

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