(2x+5)(3x+4)+(3x+2)(x+8)=(3x+2)(3x+6)+50

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Solution for (2x+5)(3x+4)+(3x+2)(x+8)=(3x+2)(3x+6)+50 equation: 



(2x+5)(3x+4)+(3x+2)(x+8)=(3x+2)(3x+6)+50

We move all terms to the left:

(2x+5)(3x+4)+(3x+2)(x+8)-((3x+2)(3x+6)+50)=0

We multiply parentheses ..

(+6x^2+8x+15x+20)+(3x+2)(x+8)-((3x+2)(3x+6)+50)=0



We calculate terms in parentheses: -((3x+2)(3x+6)+50), so:

(3x+2)(3x+6)+50

We multiply parentheses ..

(+9x^2+18x+6x+12)+50

We get rid of parentheses

9x^2+18x+6x+12+50

We add all the numbers together, and all the variables

9x^2+24x+62

Back to the equation:

-(9x^2+24x+62)



We get rid of parentheses

6x^2-9x^2+8x+15x+(3x+2)(x+8)-24x+20-62=0

We multiply parentheses ..

6x^2-9x^2+(+3x^2+24x+2x+16)+8x+15x-24x+20-62=0

We add all the numbers together, and all the variables

-3x^2+(+3x^2+24x+2x+16)-1x-42=0

We get rid of parentheses

-3x^2+3x^2+24x+2x-1x+16-42=0

We add all the numbers together, and all the variables

25x-26=0

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

25x=26

x=26/25

x=1+1/25

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