(z/5)-4=2+(3/4)

Solution for (z/5)-4=2+(3/4) equation:

(z/5)-4=2+(3/4)

We move all terms to the left:

(z/5)-4-(2+(3/4))=0

We add all the numbers together, and all the variables

(+z/5)-4-(2+(+3/4))=0

We get rid of parentheses

z/5-4-(2+(+3/4))=0

We calculate fractions


4z^2/()-4+()/()=0

We add all the numbers together, and all the variables

4z^2/()-3=0

We multiply all the terms by the denominator


4z^2-3*()=0

We add all the numbers together, and all the variables

4z^2=0

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

                              Stosujemy wzór:
$z=\frac{-b}{2a}=\frac{0}{8}=0$