9^2-x=(1/3)^3x+1

Solution for 9^2-x=(1/3)^3x+1 equation:

9^2-x=(1/3)^3x+1

We move all terms to the left:

9^2-x-((1/3)^3x+1)=0


Domain of the equation: 3)^3x+1)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

-x-((+1/3)^3x+1)+9^2=0

We add all the numbers together, and all the variables

-1x-((+1/3)^3x+1)+81=0

We multiply all the terms by the denominator


-1x*3)^3x+1)-((+81*3)^3x+1)+1=0

We add all the numbers together, and all the variables

-1x*3)^3x+1)-(243^3x+1)+1=0

Wy multiply elements

-3x^2=0

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