3y2-40y+40=0

Solution for 3y2-40y+40=0 equation:

3y^2-40y+40=0

a = 3; b = -40; c = +40;
Δ = b2-4ac
Δ = -402-4·3·40
Δ = 1120
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:

$\sqrt{\Delta}=\sqrt{1120}=\sqrt{16*70}=\sqrt{16}*\sqrt{70}=4\sqrt{70}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-4\sqrt{70}}{2*3}=\frac{40-4\sqrt{70}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+4\sqrt{70}}{2*3}=\frac{40+4\sqrt{70}}{6}
$