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

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

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

We move all terms to the left:

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

We get rid of parentheses

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


We calculate terms in parentheses: -(x+2+(-1)+(-x^2)), so:

x+2+(-1)+(-x^2)

determiningTheFunctionDomain
(-x^2)+x+2+(-1)

We add all the numbers together, and all the variables

(-x^2)+x+1

We get rid of parentheses

-x^2+x+1

We add all the numbers together, and all the variables

-1x^2+x+1

Back to the equation:

-(-1x^2+x+1)


We get rid of parentheses

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

We add all the numbers together, and all the variables

2x^2-3x+1=0

a = 2; b = -3; c = +1;
Δ = b2-4ac
Δ = -32-4·2·1
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-1}{2*2}=\frac{2}{4}
=1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+1}{2*2}=\frac{4}{4}
=1
$