3y2+32y+12=0

Solution for 3y2+32y+12=0 equation:

3y^2+32y+12=0

a = 3; b = 32; c = +12;
Δ = b2-4ac
Δ = 322-4·3·12
Δ = 880
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{880}=\sqrt{16*55}=\sqrt{16}*\sqrt{55}=4\sqrt{55}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-4\sqrt{55}}{2*3}=\frac{-32-4\sqrt{55}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+4\sqrt{55}}{2*3}=\frac{-32+4\sqrt{55}}{6}
$