(2y+12)(3y+11)=0

Solution for (2y+12)(3y+11)=0 equation:

(2y+12)(3y+11)=0

We multiply parentheses ..

(+6y^2+22y+36y+132)=0

We get rid of parentheses

6y^2+22y+36y+132=0

We add all the numbers together, and all the variables

6y^2+58y+132=0

a = 6; b = 58; c = +132;
Δ = b2-4ac
Δ = 582-4·6·132
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(58)-14}{2*6}=\frac{-72}{12}
=-6
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(58)+14}{2*6}=\frac{-44}{12}
=-3+2/3
$