3y2-4y+1=0

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Solution for 3y2-4y+1=0 equation: 



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{2}{6}
=1/3
$


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

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