S=-0.3t^2+18t

Solution for S=-0.3t^2+18t equation:

=-0.3S^2+18S

We move all terms to the left:

-(-0.3S^2+18S)=0

We get rid of parentheses

0.3S^2-18S=0

a = 0.3; b = -18; c = 0;
Δ = b2-4ac
Δ = -182-4·0.3·0
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-18}{2*0.3}=\frac{0}{0.6}
=0
$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+18}{2*0.3}=\frac{36}{0.6}
=60
$