S(t)=2+4t-3t^2

Solution for S(t)=2+4t-3t^2 equation:

(S)=2+4S-3S^2

We move all terms to the left:

(S)-(2+4S-3S^2)=0

We get rid of parentheses

3S^2-4S+S-2=0

We add all the numbers together, and all the variables

3S^2-3S-2=0

a = 3; b = -3; c = -2;
Δ = b2-4ac
Δ = -32-4·3·(-2)
Δ = 33
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{-(-3)-\sqrt{33}}{2*3}=\frac{3-\sqrt{33}}{6}
$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{33}}{2*3}=\frac{3+\sqrt{33}}{6}
$