z^2+(14/5)z+(49/25)=(64/25)

Solution for z^2+(14/5)z+(49/25)=(64/25) equation:

z^2+(14/5)z+(49/25)=(64/25)

We move all terms to the left:

z^2+(14/5)z+(49/25)-((64/25))=0


Domain of the equation: 5)z!=0

z!=0/1

z!=0

z∈R


We add all the numbers together, and all the variables

z^2+(+14/5)z+(+49/25)-((+64/25))=0

We multiply parentheses

z^2+14z^2+(+49/25)-((+64/25))=0

We get rid of parentheses

z^2+14z^2+49/25-((+64/25))=0

We calculate fractions


z^2+14z^2=0

We add all the numbers together, and all the variables

15z^2=0

a = 15; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·15·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$z=\frac{-b}{2a}=\frac{0}{30}=0$