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

Solution for (y+3)(-4y-2)=0 equation:

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

We multiply parentheses ..

(-4y^2-2y-12y-6)=0

We get rid of parentheses

-4y^2-2y-12y-6=0

We add all the numbers together, and all the variables

-4y^2-14y-6=0

a = -4; b = -14; c = -6;
Δ = b2-4ac
Δ = -142-4·(-4)·(-6)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-10}{2*-4}=\frac{4}{-8}
=-1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+10}{2*-4}=\frac{24}{-8}
=-3
$