Y=2x2-4x+1

Solution for Y=2x2-4x+1 equation:

=2Y^2-4Y+1

We move all terms to the left:

-(2Y^2-4Y+1)=0

We get rid of parentheses

-2Y^2+4Y-1=0

a = -2; b = 4; c = -1;
Δ = b2-4ac
Δ = 42-4·(-2)·(-1)
Δ = 8
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{2}}{2*-2}=\frac{-4-2\sqrt{2}}{-4}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{2}}{2*-2}=\frac{-4+2\sqrt{2}}{-4}
$