y^2-8y+6=0

Solution for y^2-8y+6=0 equation:

y^2-8y+6=0

a = 1; b = -8; c = +6;
Δ = b2-4ac
Δ = -82-4·1·6
Δ = 40
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:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{10}}{2*1}=\frac{8-2\sqrt{10}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{10}}{2*1}=\frac{8+2\sqrt{10}}{2}
$