3y^2+18y+12=0

Solution for 3y^2+18y+12=0 equation:

3y^2+18y+12=0

a = 3; b = 18; c = +12;
Δ = b2-4ac
Δ = 182-4·3·12
Δ = 180
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:

$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6\sqrt{5}}{2*3}=\frac{-18-6\sqrt{5}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6\sqrt{5}}{2*3}=\frac{-18+6\sqrt{5}}{6}
$