(3x-25)+(2x+15)(3x-20)=(6x+30)(x+20)-167x

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Solution for (3x-25)+(2x+15)(3x-20)=(6x+30)(x+20)-167x equation: 



(3x-25)+(2x+15)(3x-20)=(6x+30)(x+20)-167x

We move all terms to the left:

(3x-25)+(2x+15)(3x-20)-((6x+30)(x+20)-167x)=0

We get rid of parentheses

3x+(2x+15)(3x-20)-((6x+30)(x+20)-167x)-25=0

We multiply parentheses ..

(+6x^2-40x+45x-300)+3x-((6x+30)(x+20)-167x)-25=0



We calculate terms in parentheses: -((6x+30)(x+20)-167x), so:

(6x+30)(x+20)-167x

We add all the numbers together, and all the variables

-167x+(6x+30)(x+20)

We multiply parentheses ..

(+6x^2+120x+30x+600)-167x

We get rid of parentheses

6x^2+120x+30x-167x+600

We add all the numbers together, and all the variables

6x^2-17x+600

Back to the equation:

-(6x^2-17x+600)



We get rid of parentheses

6x^2-6x^2-40x+45x+3x+17x-300-600-25=0

We add all the numbers together, and all the variables

25x-925=0

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

25x=925

x=925/25

x=37

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