3y^2+6y=-2

Solution for 3y^2+6y=-2 equation:

3y^2+6y=-2

We move all terms to the left:

3y^2+6y-(-2)=0

We add all the numbers together, and all the variables

3y^2+6y+2=0

a = 3; b = 6; c = +2;
Δ = b2-4ac
Δ = 62-4·3·2
Δ = 12
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:

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