(3x-2)5+6x+(x-3)=(2+x)2

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Solution for (3x-2)5+6x+(x-3)=(2+x)2 equation: 



(3x-2)5+6x+(x-3)=(2+x)2

We move all terms to the left:

(3x-2)5+6x+(x-3)-((2+x)2)=0

We add all the numbers together, and all the variables

(3x-2)5+6x+(x-3)-((x+2)2)=0

We add all the numbers together, and all the variables

6x+(3x-2)5+(x-3)-((x+2)2)=0

We multiply parentheses

6x+15x+(x-3)-((x+2)2)-10=0

We get rid of parentheses

6x+15x+x-((x+2)2)-3-10=0



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

(x+2)2

We multiply parentheses

2x+4

Back to the equation:

-(2x+4)



We add all the numbers together, and all the variables

22x-(2x+4)-13=0

We get rid of parentheses

22x-2x-4-13=0

We add all the numbers together, and all the variables

20x-17=0

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

20x=17

x=17/20

x=17/20

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