3(s+2)1s=38

Solution for 3(s+2)1s=38 equation:

3(s+2)1s=38

We move all terms to the left:

3(s+2)1s-(38)=0

We multiply parentheses

3s^2+6s-38=0

a = 3; b = 6; c = -38;
Δ = b2-4ac
Δ = 62-4·3·(-38)
Δ = 492
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:

$\sqrt{\Delta}=\sqrt{492}=\sqrt{4*123}=\sqrt{4}*\sqrt{123}=2\sqrt{123}$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{123}}{2*3}=\frac{-6-2\sqrt{123}}{6}
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{123}}{2*3}=\frac{-6+2\sqrt{123}}{6}
$