c=1/2+(c1-c2)

Solution for c=1/2+(c1-c2) equation:

c=1/2+(c1-c2)

We move all terms to the left:

c-(1/2+(c1-c2))=0


Domain of the equation: 2+(c1-c2))!=0

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

(c1-c2))!=-2

c∈R


We add all the numbers together, and all the variables

-(1/2+(+c-1c^2))+c=0

We multiply all the terms by the denominator


-(1+c*2+(+c-1c^2))=0


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

1+c*2+(+c-1c^2)

determiningTheFunctionDomain
(+c-1c^2)+c*2+1

Wy multiply elements

(+c-1c^2)+2c+1

We get rid of parentheses

-1c^2+c+2c+1

We add all the numbers together, and all the variables

-1c^2+3c+1

Back to the equation:

-(-1c^2+3c+1)


We get rid of parentheses

1c^2-3c-1=0

We add all the numbers together, and all the variables

c^2-3c-1=0

a = 1; b = -3; c = -1;
Δ = b2-4ac
Δ = -32-4·1·(-1)
Δ = 13
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{13}}{2*1}=\frac{3-\sqrt{13}}{2}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{13}}{2*1}=\frac{3+\sqrt{13}}{2}
$