2z(z+1)+3(z+2)=3z(z+2)-z2

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Solution for 2z(z+1)+3(z+2)=3z(z+2)-z2 equation: 



2z(z+1)+3(z+2)=3z(z+2)-z2

We move all terms to the left:

2z(z+1)+3(z+2)-(3z(z+2)-z2)=0

We multiply parentheses

2z^2+2z+3z-(3z(z+2)-z2)+6=0



We calculate terms in parentheses: -(3z(z+2)-z2), so:

3z(z+2)-z2

We add all the numbers together, and all the variables

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

We multiply parentheses

-1z^2+3z^2+6z

We add all the numbers together, and all the variables

2z^2+6z

Back to the equation:

-(2z^2+6z)



We add all the numbers together, and all the variables

2z^2+5z-(2z^2+6z)+6=0

We get rid of parentheses

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

We add all the numbers together, and all the variables

-1z+6=0

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

-z=-6

z=-6/-1

z=+6

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