-4y^2-8y+12=0

Solution for -4y^2-8y+12=0 equation:

-4y^2-8y+12=0

a = -4; b = -8; c = +12;
Δ = b2-4ac
Δ = -82-4·(-4)·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{-(-8)-16}{2*-4}=\frac{-8}{-8}
=1
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+16}{2*-4}=\frac{24}{-8}
=-3
$