3y^2-33y+39=0

Solution for 3y^2-33y+39=0 equation:

3y^2-33y+39=0

a = 3; b = -33; c = +39;
Δ = b2-4ac
Δ = -332-4·3·39
Δ = 621
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:

$\sqrt{\Delta}=\sqrt{621}=\sqrt{9*69}=\sqrt{9}*\sqrt{69}=3\sqrt{69}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-33)-3\sqrt{69}}{2*3}=\frac{33-3\sqrt{69}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-33)+3\sqrt{69}}{2*3}=\frac{33+3\sqrt{69}}{6}
$