3s2+12s+8=0

Solution for 3s2+12s+8=0 equation:

3s^2+12s+8=0

a = 3; b = 12; c = +8;
Δ = b2-4ac
Δ = 122-4·3·8
Δ = 48
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{3}}{2*3}=\frac{-12-4\sqrt{3}}{6}
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{3}}{2*3}=\frac{-12+4\sqrt{3}}{6}
$