16y^2+12=-32y

Solution for 16y^2+12=-32y equation:

16y^2+12=-32y

We move all terms to the left:

16y^2+12-(-32y)=0

We get rid of parentheses

16y^2+32y+12=0

a = 16; b = 32; c = +12;
Δ = b2-4ac
Δ = 322-4·16·12
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{-(32)-16}{2*16}=\frac{-48}{32}
=-1+1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+16}{2*16}=\frac{-16}{32}
=-1/2
$