-4.9t^2+54t-0.3=0

Solution for -4.9t^2+54t-0.3=0 equation:

-4.9t^2+54t-0.3=0

a = -4.9; b = 54; c = -0.3;
Δ = b2-4ac
Δ = 542-4·(-4.9)·(-0.3)
Δ = 2910.12
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(54)-\sqrt{2910.12}}{2*-4.9}=\frac{-54-\sqrt{2910.12}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(54)+\sqrt{2910.12}}{2*-4.9}=\frac{-54+\sqrt{2910.12}}{-9.8}
$