x-x(3x+5)=x-2(x2+x)-x2-14

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Solution for x-x(3x+5)=x-2(x2+x)-x2-14 equation: 



x-x(3x+5)=x-2(x2+x)-x2-14

We move all terms to the left:

x-x(3x+5)-(x-2(x2+x)-x2-14)=0

We add all the numbers together, and all the variables

-(x-2(+x^2+x)-x2-14)+x-x(3x+5)=0

We multiply parentheses

-(x-2(+x^2+x)-x2-14)-3x^2+x-5x=0



We calculate terms in parentheses: -(x-2(+x^2+x)-x2-14), so:

x-2(+x^2+x)-x2-14

determiningTheFunctionDomain
-2(+x^2+x)+x-x2-14

We add all the numbers together, and all the variables

-1x^2-2(+x^2+x)+x-14

We multiply parentheses

-1x^2-2x^2-2x+x-14

We add all the numbers together, and all the variables

-3x^2-1x-14

Back to the equation:

-(-3x^2-1x-14)



We add all the numbers together, and all the variables

-3x^2-(-3x^2-1x-14)-4x=0

We get rid of parentheses

-3x^2+3x^2+1x-4x+14=0

We add all the numbers together, and all the variables

-3x+14=0

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

-3x=-14

x=-14/-3

x=4+2/3

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