-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{-24}{-8}
=+3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+16}{2*-4}=\frac{8}{-8}
=-1
$