0=-20t^2+27t

Solution for 0=-20t^2+27t equation:

0=-20t^2+27t

We move all terms to the left:

0-(-20t^2+27t)=0

We add all the numbers together, and all the variables

-(-20t^2+27t)=0

We get rid of parentheses

20t^2-27t=0

a = 20; b = -27; c = 0;
Δ = b2-4ac
Δ = -272-4·20·0
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*20}=\frac{0}{40}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+27}{2*20}=\frac{54}{40}
=1+7/20
$