k+4=(7k-4)/(3)

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Solution for k+4=(7k-4)/(3) equation: 



k+4=(7k-4)/(3)

We move all terms to the left:

k+4-((7k-4)/(3))=0

We multiply all the terms by the denominator


k*3)-((7k-4)+4*3)=0

We add all the numbers together, and all the variables

k*3)-((7k-4)=0

Wy multiply elements

3k^2=0

a = 3; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·3·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$k=\frac{-b}{2a}=\frac{0}{6}=0$

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