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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+3y^2+9y+y+3)=0

We get rid of parentheses

3y^2+9y+y+3=0

We add all the numbers together, and all the variables

3y^2+10y+3=0

a = 3; b = 10; c = +3;
Δ = b2-4ac
Δ = 102-4·3·3
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-8}{2*3}=\frac{-18}{6}
=-3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+8}{2*3}=\frac{-2}{6}
=-1/3
$