18x+4x(2x-4)=2(4x2-x)+20x

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Solution for 18x+4x(2x-4)=2(4x2-x)+20x equation: 



18x+4x(2x-4)=2(4x^2-x)+20x

We move all terms to the left:

18x+4x(2x-4)-(2(4x^2-x)+20x)=0

We multiply parentheses

8x^2+18x-16x-(2(4x^2-x)+20x)=0



We calculate terms in parentheses: -(2(4x^2-x)+20x), so:

2(4x^2-x)+20x

We add all the numbers together, and all the variables

20x+2(4x^2-x)

We multiply parentheses

8x^2+20x-2x

We add all the numbers together, and all the variables

8x^2+18x

Back to the equation:

-(8x^2+18x)



We add all the numbers together, and all the variables

8x^2+2x-(8x^2+18x)=0

We get rid of parentheses

8x^2-8x^2+2x-18x=0

We add all the numbers together, and all the variables

-16x=0

x=0/-16

x=0

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