4r-4=(r-4)/5

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Solution for 4r-4=(r-4)/5 equation: 



4r-4=(r-4)/5

We move all terms to the left:

4r-4-((r-4)/5)=0

We multiply all the terms by the denominator


4r*5)-((r-4)-4*5)=0

We add all the numbers together, and all the variables

4r*5)-((r-4)=0

Wy multiply elements

20r^2=0

a = 20; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·20·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$r=\frac{-b}{2a}=\frac{0}{40}=0$

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