2x(3-x)-4(x+2)=3x(2-3x)+5+7x2-12

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Solution for 2x(3-x)-4(x+2)=3x(2-3x)+5+7x2-12 equation: 



2x(3-x)-4(x+2)=3x(2-3x)+5+7x^2-12

We move all terms to the left:

2x(3-x)-4(x+2)-(3x(2-3x)+5+7x^2-12)=0

We add all the numbers together, and all the variables

-(3x(-3x+2)+5+7x^2-12)+2x(-1x+3)-4(x+2)=0

We multiply parentheses

-(3x(-3x+2)+5+7x^2-12)-2x^2+6x-4x-8=0



We calculate terms in parentheses: -(3x(-3x+2)+5+7x^2-12), so:

3x(-3x+2)+5+7x^2-12

determiningTheFunctionDomain
7x^2+3x(-3x+2)+5-12

We add all the numbers together, and all the variables

7x^2+3x(-3x+2)-7

We multiply parentheses

7x^2-9x^2+6x-7

We add all the numbers together, and all the variables

-2x^2+6x-7

Back to the equation:

-(-2x^2+6x-7)



We add all the numbers together, and all the variables

-2x^2-(-2x^2+6x-7)+2x-8=0

We get rid of parentheses

-2x^2+2x^2-6x+2x+7-8=0

We add all the numbers together, and all the variables

-4x-1=0

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

-4x=1

x=1/-4

x=-1/4

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