8y^2-48y+32=0

Solution for 8y^2-48y+32=0 equation:

8y^2-48y+32=0

a = 8; b = -48; c = +32;
Δ = b2-4ac
Δ = -482-4·8·32
Δ = 1280
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:

$\sqrt{\Delta}=\sqrt{1280}=\sqrt{256*5}=\sqrt{256}*\sqrt{5}=16\sqrt{5}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-16\sqrt{5}}{2*8}=\frac{48-16\sqrt{5}}{16}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+16\sqrt{5}}{2*8}=\frac{48+16\sqrt{5}}{16}
$