202.5=18t-0.2t^2

Solution for 202.5=18t-0.2t^2 equation:

202.5=18t-0.2t^2

We move all terms to the left:

202.5-(18t-0.2t^2)=0

We get rid of parentheses

0.2t^2-18t+202.5=0

a = 0.2; b = -18; c = +202.5;
Δ = b2-4ac
Δ = -182-4·0.2·202.5
Δ = 162
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{162}=\sqrt{81*2}=\sqrt{81}*\sqrt{2}=9\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-9\sqrt{2}}{2*0.2}=\frac{18-9\sqrt{2}}{0.4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+9\sqrt{2}}{2*0.2}=\frac{18+9\sqrt{2}}{0.4}
$