-0.3t^2+27t=0

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

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