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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+3y^2+2y+18y+12)=0

We get rid of parentheses

3y^2+2y+18y+12=0

We add all the numbers together, and all the variables

3y^2+20y+12=0

a = 3; b = 20; c = +12;
Δ = b2-4ac
Δ = 202-4·3·12
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*3}=\frac{-36}{6}
=-6
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+16}{2*3}=\frac{-4}{6}
=-2/3
$