S(t)=16t^2+68t

Solution for S(t)=16t^2+68t equation:

(S)=16S^2+68S

We move all terms to the left:

(S)-(16S^2+68S)=0

We get rid of parentheses

-16S^2+S-68S=0

We add all the numbers together, and all the variables

-16S^2-67S=0

a = -16; b = -67; c = 0;
Δ = b2-4ac
Δ = -672-4·(-16)·0
Δ = 4489
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4489}=67$
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-67)-67}{2*-16}=\frac{0}{-32}
=0
$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-67)+67}{2*-16}=\frac{134}{-32}
=-4+3/16
$