(3x+1)(x-5)=(x-2)(3x-3)+4

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Solution for (3x+1)(x-5)=(x-2)(3x-3)+4 equation: 



(3x+1)(x-5)=(x-2)(3x-3)+4

We move all terms to the left:

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

We multiply parentheses ..

(+3x^2-15x+x-5)-((x-2)(3x-3)+4)=0



We calculate terms in parentheses: -((x-2)(3x-3)+4), so:

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

We multiply parentheses ..

(+3x^2-3x-6x+6)+4

We get rid of parentheses

3x^2-3x-6x+6+4

We add all the numbers together, and all the variables

3x^2-9x+10

Back to the equation:

-(3x^2-9x+10)



We get rid of parentheses

3x^2-3x^2-15x+x+9x-5-10=0

We add all the numbers together, and all the variables

-5x-15=0

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

-5x=15

x=15/-5

x=-3

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