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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-3S^2+5S-2=0

a = -3; b = 5; c = -2;
Δ = b2-4ac
Δ = 52-4·(-3)·(-2)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-1}{2*-3}=\frac{-6}{-6}
=1
$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+1}{2*-3}=\frac{-4}{-6}
=2/3
$