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

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



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

We move all terms to the left:

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

We multiply parentheses ..

(+12x^2+18x-10x-15)-((3x+8)(4x-2))=0



We calculate terms in parentheses: -((3x+8)(4x-2)), so:

(3x+8)(4x-2)

We multiply parentheses ..

(+12x^2-6x+32x-16)

We get rid of parentheses

12x^2-6x+32x-16

We add all the numbers together, and all the variables

12x^2+26x-16

Back to the equation:

-(12x^2+26x-16)



We get rid of parentheses

12x^2-12x^2+18x-10x-26x-15+16=0

We add all the numbers together, and all the variables

-18x+1=0

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

-18x=-1

x=-1/-18

x=1/18

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