7x(4+3x)=3x(7x+1)+100

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Solution for 7x(4+3x)=3x(7x+1)+100 equation: 



7x(4+3x)=3x(7x+1)+100

We move all terms to the left:

7x(4+3x)-(3x(7x+1)+100)=0

We add all the numbers together, and all the variables

7x(3x+4)-(3x(7x+1)+100)=0

We multiply parentheses

21x^2+28x-(3x(7x+1)+100)=0



We calculate terms in parentheses: -(3x(7x+1)+100), so:

3x(7x+1)+100

We multiply parentheses

21x^2+3x+100

Back to the equation:

-(21x^2+3x+100)



We get rid of parentheses

21x^2-21x^2+28x-3x-100=0

We add all the numbers together, and all the variables

25x-100=0

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

25x=100

x=100/25

x=4

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