(3/5)(2x-4)=18

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Solution for (3/5)(2x-4)=18 equation: 



(3/5)(2x-4)=18

We move all terms to the left:

(3/5)(2x-4)-(18)=0



Domain of the equation: 5)(2x-4)!=0

x∈R



We add all the numbers together, and all the variables

(+3/5)(2x-4)-18=0

We multiply parentheses ..

(+6x^2+3/5*-4)-18=0

We multiply all the terms by the denominator


(+6x^2+3-18*5*-4)=0

We get rid of parentheses

6x^2+3-4-18*5*=0

We add all the numbers together, and all the variables

6x^2=0

a = 6; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·6·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{0}{12}=0$

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