3(z/1)+11=-2(z/13)

Solution for 3(z/1)+11=-2(z/13) equation:

3(z/1)+11=-2(z/13)

We move all terms to the left:

3(z/1)+11-(-2(z/13))=0

We add all the numbers together, and all the variables

3(+z/1)-(-2(+z/13))+11=0

We multiply parentheses

3z-(-2(+z/13))+11=0

We multiply all the terms by the denominator


3z*13))-(-2(+z+11*13))=0

We add all the numbers together, and all the variables

3z*13))-(-2(z+143))=0

We add all the numbers together, and all the variables

3z*13))-(-2(z=0

Wy multiply elements

39z^2=0

a = 39; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·39·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$z=\frac{-b}{2a}=\frac{0}{78}=0$