0=-0.2t^2+18t

Solution for 0=-0.2t^2+18t equation:

0=-0.2t^2+18t

We move all terms to the left:

0-(-0.2t^2+18t)=0

We add all the numbers together, and all the variables

-(-0.2t^2+18t)=0

We get rid of parentheses

0.2t^2-18t=0

a = 0.2; b = -18; c = 0;
Δ = b2-4ac
Δ = -182-4·0.2·0
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-18}{2*0.2}=\frac{0}{0.4}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+18}{2*0.2}=\frac{36}{0.4}
=90
$