3x-7/12=1/6*(2x-3)=3-x/2-1/2

Solution for 3x-7/12=1/6*(2x-3)=3-x/2-1/2 equation:

3x-7/12=1/6(2x-3)=3-x/2-1/2

We move all terms to the left:

3x-7/12-(1/6(2x-3))=0


Domain of the equation: 6(2x-3))!=0

x∈R


We calculate fractions


3x+()/144x+(-42x2/144x=0

We multiply all the terms by the denominator


3x*144x+(-42x2+()=0


We calculate terms in parentheses: +(-42x2+(), so:

-42x2+(

We add all the numbers together, and all the variables

-42x^2

Back to the equation:

+(-42x^2)


Wy multiply elements

(-42x^2)+432x^2=0

We get rid of parentheses

-42x^2+432x^2=0

We add all the numbers together, and all the variables

390x^2=0

a = 390; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·390·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{0}{780}=0$