S=12t+-2t2

Solution for S=12t+-2t2 equation:

=12S+-2S^2

We move all terms to the left:

-(12S+-2S^2)=0

We use the square of the difference formula

-(12S-2S^2)=0

We get rid of parentheses

2S^2-12S=0

a = 2; b = -12; c = 0;
Δ = b2-4ac
Δ = -122-4·2·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-12}{2*2}=\frac{0}{4}
=0
$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+12}{2*2}=\frac{24}{4}
=6
$