(c-3)=(c2-2c+3)

Solution for (c-3)=(c2-2c+3) equation:

(c-3)=(c2-2c+3)

We move all terms to the left:

(c-3)-((c2-2c+3))=0

We add all the numbers together, and all the variables

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

We get rid of parentheses

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


We calculate terms in parentheses: -((+c^2-2c+3)), so:

(+c^2-2c+3)

We get rid of parentheses

c^2-2c+3

Back to the equation:

-(c^2-2c+3)


We add all the numbers together, and all the variables

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

We get rid of parentheses

-c^2+c+2c-3-3=0

We add all the numbers together, and all the variables

-1c^2+3c-6=0

a = -1; b = 3; c = -6;
Δ = b2-4ac
Δ = 32-4·(-1)·(-6)
Δ = -15
Delta is less than zero, so there is no solution for the equation