-(3s^2+12s+8)=0

Solution for -(3s^2+12s+8)=0 equation:

-(3s^2+12s+8)=0

We get rid of parentheses

-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}
$