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

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

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

We move all terms to the left:

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


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

2(x^2-x+1)

We multiply parentheses

2x^2-2x+2

Back to the equation:

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


We get rid of parentheses

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

We add all the numbers together, and all the variables

-1x^2+1=0

a = -1; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-1)·1
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2}{2*-1}=\frac{-2}{-2}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2}{2*-1}=\frac{2}{-2}
=-1
$