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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+6y^2+4y+3y+2)-2=0

We get rid of parentheses

6y^2+4y+3y+2-2=0

We add all the numbers together, and all the variables

6y^2+7y=0

a = 6; b = 7; c = 0;
Δ = b2-4ac
Δ = 72-4·6·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-7}{2*6}=\frac{-14}{12}
=-1+1/6
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+7}{2*6}=\frac{0}{12}
=0
$