x2=2(x2-3x+1)

Solution for x2=2(x2-3x+1) equation:

x2=2(x2-3x+1)

We move all terms to the left:

x2-(2(x2-3x+1))=0

We add all the numbers together, and all the variables

-(2(+x^2-3x+1))+x2=0

We add all the numbers together, and all the variables

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


We calculate terms in parentheses: -(2(+x^2-3x+1)), so:

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

We multiply parentheses

2x^2-6x+2

Back to the equation:

-(2x^2-6x+2)


We get rid of parentheses

x^2-2x^2+6x-2=0

We add all the numbers together, and all the variables

-1x^2+6x-2=0

a = -1; b = 6; c = -2;
Δ = b2-4ac
Δ = 62-4·(-1)·(-2)
Δ = 28
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{7}}{2*-1}=\frac{-6-2\sqrt{7}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{7}}{2*-1}=\frac{-6+2\sqrt{7}}{-2}
$