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

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

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

We move all terms to the left:

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


Domain of the equation: 2)x^2-1)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

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

We multiply all the terms by the denominator


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

We add all the numbers together, and all the variables

-1x*2)x^2-1)-(6x^2-1)+1=0

Wy multiply elements

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