3y^2-12y-2=0

Solution for 3y^2-12y-2=0 equation:

3y^2-12y-2=0

a = 3; b = -12; c = -2;
Δ = b2-4ac
Δ = -122-4·3·(-2)
Δ = 168
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:

$\sqrt{\Delta}=\sqrt{168}=\sqrt{4*42}=\sqrt{4}*\sqrt{42}=2\sqrt{42}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{42}}{2*3}=\frac{12-2\sqrt{42}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{42}}{2*3}=\frac{12+2\sqrt{42}}{6}
$