-12y^2+8y+32=0

Solution for -12y^2+8y+32=0 equation:

-12y^2+8y+32=0

a = -12; b = 8; c = +32;
Δ = b2-4ac
Δ = 82-4·(-12)·32
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-40}{2*-12}=\frac{-48}{-24}
=+2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+40}{2*-12}=\frac{32}{-24}
=-1+1/3
$