3y2-20y+32=0

Solution for 3y2-20y+32=0 equation:

3y^2-20y+32=0

a = 3; b = -20; c = +32;
Δ = b2-4ac
Δ = -202-4·3·32
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4}{2*3}=\frac{16}{6}
=2+2/3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4}{2*3}=\frac{24}{6}
=4
$