7=(1/4)(3x-12)

Solution for 7=(1/4)(3x-12) equation:

7=(1/4)(3x-12)

We move all terms to the left:

7-((1/4)(3x-12))=0


Domain of the equation: 4)(3x-12))!=0

x∈R


We add all the numbers together, and all the variables

-((+1/4)(3x-12))+7=0

We multiply parentheses ..

-((+3x^2+1/4*-12))+7=0

We multiply all the terms by the denominator


-((+3x^2+1+7*4*-12))=0


We calculate terms in parentheses: -((+3x^2+1+7*4*-12)), so:

(+3x^2+1+7*4*-12)

We get rid of parentheses

3x^2+1-12+7*4*

We add all the numbers together, and all the variables

3x^2

Back to the equation:

-(3x^2)


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:
$x=\frac{-b}{2a}=\frac{0}{-6}=0$