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x(x+4)-2(3x-1)=3x(1-x)+x(4x-5)
We move all terms to the left:
x(x+4)-2(3x-1)-(3x(1-x)+x(4x-5))=0
We add all the numbers together, and all the variables
x(x+4)-2(3x-1)-(3x(-1x+1)+x(4x-5))=0
We multiply parentheses
x^2+4x-6x-(3x(-1x+1)+x(4x-5))+2=0
We calculate terms in parentheses: -(3x(-1x+1)+x(4x-5)), so:We add all the numbers together, and all the variables
3x(-1x+1)+x(4x-5)
We multiply parentheses
-3x^2+4x^2+3x-5x
We add all the numbers together, and all the variables
x^2-2x
Back to the equation:
-(x^2-2x)
x^2-2x-(x^2-2x)+2=0
We get rid of parentheses
x^2-x^2-2x+2x+2=0
We add all the numbers together, and all the variables
2!=0
There is no solution for this equation
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