4y^2-8y+1=0

Solution for 4y^2-8y+1=0 equation:

4y^2-8y+1=0

a = 4; b = -8; c = +1;
Δ = b2-4ac
Δ = -82-4·4·1
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4\sqrt{3}}{2*4}=\frac{8-4\sqrt{3}}{8}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4\sqrt{3}}{2*4}=\frac{8+4\sqrt{3}}{8}
$