Y=3x2-4x+1

Solution for Y=3x2-4x+1 equation:

=3Y^2-4Y+1

We move all terms to the left:

-(3Y^2-4Y+1)=0

We get rid of parentheses

-3Y^2+4Y-1=0

a = -3; b = 4; c = -1;
Δ = b2-4ac
Δ = 42-4·(-3)·(-1)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2}{2*-3}=\frac{-6}{-6}
=1
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2}{2*-3}=\frac{-2}{-6}
=1/3
$