-4.9t^2+27t=0

Solution for -4.9t^2+27t=0 equation:

-4.9t^2+27t=0

a = -4.9; b = 27; c = 0;
Δ = b2-4ac
Δ = 272-4·(-4.9)·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*-4.9}=\frac{-54}{-9.8}
=5+5/9.8
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+27}{2*-4.9}=\frac{0}{-9.8}
=0
$