5(r+3)=6(2r-4)/3

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Solution for 5(r+3)=6(2r-4)/3 equation: 



5(r+3)=6(2r-4)/3

We move all terms to the left:

5(r+3)-(6(2r-4)/3)=0

We multiply parentheses

5r-(6(2r-4)/3)+15=0

We multiply all the terms by the denominator


5r*3)-(6(2r-4)+15*3)=0

We add all the numbers together, and all the variables

5r*3)-(6(2r-4)=0

Wy multiply elements

15r^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:
$r=\frac{-b}{2a}=\frac{0}{30}=0$

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