45=(1/3)(3k-12)

Solution for 45=(1/3)(3k-12) equation:

45=(1/3)(3k-12)

We move all terms to the left:

45-((1/3)(3k-12))=0


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

k∈R


We add all the numbers together, and all the variables

-((+1/3)(3k-12))+45=0

We multiply parentheses ..

-((+3k^2+1/3*-12))+45=0

We multiply all the terms by the denominator


-((+3k^2+1+45*3*-12))=0


We calculate terms in parentheses: -((+3k^2+1+45*3*-12)), so:

(+3k^2+1+45*3*-12)

We get rid of parentheses

3k^2+1-12+45*3*

We add all the numbers together, and all the variables

3k^2

Back to the equation:

-(3k^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:
$k=\frac{-b}{2a}=\frac{0}{-6}=0$