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

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

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

We move all terms to the left:

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


Domain of the equation: 5)(z-2)!=0

z∈R


We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+2z^2+2/5*-2)+3=0

We multiply all the terms by the denominator


(+2z^2+2+3*5*-2)=0

We get rid of parentheses

2z^2+2-2+3*5*=0

We add all the numbers together, and all the variables

2z^2=0

a = 2; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·2·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$z=\frac{-b}{2a}=\frac{0}{4}=0$