2x*(x-3)=x*(2x+4)-80

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Solution for 2x*(x-3)=x*(2x+4)-80 equation: 



2x(x-3)=x(2x+4)-80

We move all terms to the left:

2x(x-3)-(x(2x+4)-80)=0

We multiply parentheses

2x^2-6x-(x(2x+4)-80)=0



We calculate terms in parentheses: -(x(2x+4)-80), so:

x(2x+4)-80

We multiply parentheses

2x^2+4x-80

Back to the equation:

-(2x^2+4x-80)



We get rid of parentheses

2x^2-2x^2-6x-4x+80=0

We add all the numbers together, and all the variables

-10x+80=0

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

-10x=-80

x=-80/-10

x=+8

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