(x)=x^2/3+x

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

(x)=x^2/3+x

We move all terms to the left:

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


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

We move all terms containing x to the left, all other terms to the right

x)!=-3

x!=-3/1

x!=-3

x∈R


We get rid of parentheses

-x^2/3+x-x=0

We multiply all the terms by the denominator


-x^2+x*3-x*3=0

We add all the numbers together, and all the variables

-1x^2+x*3-x*3=0

Wy multiply elements

-1x^2+3x-3x=0

We add all the numbers together, and all the variables

-1x^2=0

a = -1; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·(-1)·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

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