(3y+3)(y+2)=5

Solution for (3y+3)(y+2)=5 equation:

(3y+3)(y+2)=5

We move all terms to the left:

(3y+3)(y+2)-(5)=0

We multiply parentheses ..

(+3y^2+6y+3y+6)-5=0

We get rid of parentheses

3y^2+6y+3y+6-5=0

We add all the numbers together, and all the variables

3y^2+9y+1=0

a = 3; b = 9; c = +1;
Δ = b2-4ac
Δ = 92-4·3·1
Δ = 69
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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