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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

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


We calculate terms in parentheses: -(3(z+1)2z), so:

3(z+1)2z

We multiply parentheses

6z^2+6z

Back to the equation:

-(6z^2+6z)


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-6z^2-1z-2=0

a = -6; b = -1; c = -2;
Δ = b2-4ac
Δ = -12-4·(-6)·(-2)
Δ = -47
Delta is less than zero, so there is no solution for the equation