(2z+1)(3z+2)=(3z+2)(2z-3)

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Solution for (2z+1)(3z+2)=(3z+2)(2z-3) equation: 



(2z+1)(3z+2)=(3z+2)(2z-3)

We move all terms to the left:

(2z+1)(3z+2)-((3z+2)(2z-3))=0

We multiply parentheses ..

(+6z^2+4z+3z+2)-((3z+2)(2z-3))=0



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

(3z+2)(2z-3)

We multiply parentheses ..

(+6z^2-9z+4z-6)

We get rid of parentheses

6z^2-9z+4z-6

We add all the numbers together, and all the variables

6z^2-5z-6

Back to the equation:

-(6z^2-5z-6)



We get rid of parentheses

6z^2-6z^2+4z+3z+5z+2+6=0

We add all the numbers together, and all the variables

12z+8=0

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

12z=-8

z=-8/12

z=-2/3

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