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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We multiply all the terms by the denominator


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

We add all the numbers together, and all the variables

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

Wy multiply elements

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-2x^2=0

a = -2; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·(-2)·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{0}{-4}=0$