E=I(R-r);r

Solution for E=I(R-r);r equation:

=(-E)E

We move all terms to the left:

-((-E)E)=0

We add all the numbers together, and all the variables

-((-1E)E)=0


We calculate terms in parentheses: -((-1E)E), so:

(-1E)E

We multiply parentheses

-1E^2

Back to the equation:

-(-1E^2)


We get rid of parentheses

1E^2=0

We add all the numbers together, and all the variables

E^2=0

a = 1; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·1·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

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