2y(3y+1)+6(3y+1)=0

Solution for 2y(3y+1)+6(3y+1)=0 equation:

2y(3y+1)+6(3y+1)=0

We multiply parentheses

6y^2+2y+18y+6=0

We add all the numbers together, and all the variables

6y^2+20y+6=0

a = 6; b = 20; c = +6;
Δ = b2-4ac
Δ = 202-4·6·6
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-16}{2*6}=\frac{-36}{12}
=-3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+16}{2*6}=\frac{-4}{12}
=-1/3
$