3s^2+60s+200=0

Solution for 3s^2+60s+200=0 equation:

3s^2+60s+200=0

a = 3; b = 60; c = +200;
Δ = b2-4ac
Δ = 602-4·3·200
Δ = 1200
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:

$\sqrt{\Delta}=\sqrt{1200}=\sqrt{400*3}=\sqrt{400}*\sqrt{3}=20\sqrt{3}$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-20\sqrt{3}}{2*3}=\frac{-60-20\sqrt{3}}{6}
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+20\sqrt{3}}{2*3}=\frac{-60+20\sqrt{3}}{6}
$