(3+z)2z=3z(z+2)

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

(3+z)2z=3z(z+2)

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

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


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

3z(z+2)

We multiply parentheses

3z^2+6z

Back to the equation:

-(3z^2+6z)


We get rid of parentheses

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

We add all the numbers together, and all the variables

-1z^2=0

a = -1; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·(-1)·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$z=\frac{-b}{2a}=\frac{0}{-2}=0$