s(3s+0.22)+(s-0.12)=4.6

Solution for s(3s+0.22)+(s-0.12)=4.6 equation:

s(3s+0.22)+(s-0.12)=4.6

We move all terms to the left:

s(3s+0.22)+(s-0.12)-(4.6)=0

We add all the numbers together, and all the variables

s(3s+0.22)+(s-0.12)-4.6=0

We multiply parentheses

3s^2+0.22s+(s-0.12)-4.6=0

We get rid of parentheses

3s^2+0.22s+s-0.12-4.6=0

We add all the numbers together, and all the variables

3s^2+1.22s-4.72=0

a = 3; b = 1.22; c = -4.72;
Δ = b2-4ac
Δ = 1.222-4·3·(-4.72)
Δ = 58.1284
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1.22)-\sqrt{58.1284}}{2*3}=\frac{-1.22-\sqrt{58.1284}}{6}
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.22)+\sqrt{58.1284}}{2*3}=\frac{-1.22+\sqrt{58.1284}}{6}
$