3s^2+18s+20=0

Solution for 3s^2+18s+20=0 equation:

3s^2+18s+20=0

a = 3; b = 18; c = +20;
Δ = b2-4ac
Δ = 182-4·3·20
Δ = 84
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:

$\sqrt{\Delta}=\sqrt{84}=\sqrt{4*21}=\sqrt{4}*\sqrt{21}=2\sqrt{21}$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{21}}{2*3}=\frac{-18-2\sqrt{21}}{6}
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{21}}{2*3}=\frac{-18+2\sqrt{21}}{6}
$