y^2-32y+8=0

Solution for y^2-32y+8=0 equation:

y^2-32y+8=0

a = 1; b = -32; c = +8;
Δ = b2-4ac
Δ = -322-4·1·8
Δ = 992
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:

$\sqrt{\Delta}=\sqrt{992}=\sqrt{16*62}=\sqrt{16}*\sqrt{62}=4\sqrt{62}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-4\sqrt{62}}{2*1}=\frac{32-4\sqrt{62}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+4\sqrt{62}}{2*1}=\frac{32+4\sqrt{62}}{2}
$