4x-18=((3x-4)+(x+12))/2

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Solution for 4x-18=((3x-4)+(x+12))/2 equation: 



4x-18=((3x-4)+(x+12))/2

We move all terms to the left:

4x-18-(((3x-4)+(x+12))/2)=0

We multiply all the terms by the denominator


4x*2)-(((3x-4)+(x+12))-18*2)=0

We add all the numbers together, and all the variables

4x*2)-(((3x-4)+(x+12))=0

Wy multiply elements

8x^2=0

a = 8; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·8·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{0}{16}=0$

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