3y^2-24y+32=0

Solution for 3y^2-24y+32=0 equation:

3y^2-24y+32=0

a = 3; b = -24; c = +32;
Δ = b2-4ac
Δ = -242-4·3·32
Δ = 192
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:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-8\sqrt{3}}{2*3}=\frac{24-8\sqrt{3}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+8\sqrt{3}}{2*3}=\frac{24+8\sqrt{3}}{6}
$