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

Solution for -2y^2-3y-1=0 equation:

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

a = -2; b = -3; c = -1;
Δ = b2-4ac
Δ = -32-4·(-2)·(-1)
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*-2}=\frac{2}{-4}
=-1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+1}{2*-2}=\frac{4}{-4}
=-1
$