z/5=(3/10)z+4

Solution for z/5=(3/10)z+4 equation:

z/5=(3/10)z+4

We move all terms to the left:

z/5-((3/10)z+4)=0


Domain of the equation: 10)z+4)!=0

z!=0/1

z!=0

z∈R


We add all the numbers together, and all the variables

z/5-((+3/10)z+4)=0

We calculate fractions


10z^2/50z+()/50z=0

We multiply all the terms by the denominator


10z^2+()=0

We add all the numbers together, and all the variables

10z^2=0

a = 10; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·10·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$z=\frac{-b}{2a}=\frac{0}{20}=0$