-y^2-3y-2=0

Solution for -y^2-3y-2=0 equation:

-y^2-3y-2=0

We add all the numbers together, and all the variables

-1y^2-3y-2=0

a = -1; b = -3; c = -2;
Δ = b2-4ac
Δ = -32-4·(-1)·(-2)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-1}{2*-1}=\frac{2}{-2}
=-1
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+1}{2*-1}=\frac{4}{-2}
=-2
$