S(t)=18t+4.9t^2

Solution for S(t)=18t+4.9t^2 equation:

(S)=18S+4.9S^2

We move all terms to the left:

(S)-(18S+4.9S^2)=0

We get rid of parentheses

-4.9S^2-18S+S=0

We add all the numbers together, and all the variables

-4.9S^2-17S=0

a = -4.9; b = -17; c = 0;
Δ = b2-4ac
Δ = -172-4·(-4.9)·0
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-17}{2*-4.9}=\frac{0}{-9.8}
=0
$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+17}{2*-4.9}=\frac{34}{-9.8}
=-3+2.875/6.125
$