0=-0.3t^2+27t

Solution for 0=-0.3t^2+27t equation:

0=-0.3t^2+27t

We move all terms to the left:

0-(-0.3t^2+27t)=0

We add all the numbers together, and all the variables

-(-0.3t^2+27t)=0

We get rid of parentheses

0.3t^2-27t=0

a = 0.3; b = -27; c = 0;
Δ = b2-4ac
Δ = -272-4·0.3·0
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-27)-27}{2*0.3}=\frac{0}{0.6}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+27}{2*0.3}=\frac{54}{0.6}
=90
$