-4.8t^2+19.6t+58.8=0

Solution for -4.8t^2+19.6t+58.8=0 equation:

-4.8t^2+19.6t+58.8=0

a = -4.8; b = 19.6; c = +58.8;
Δ = b2-4ac
Δ = 19.62-4·(-4.8)·58.8
Δ = 1513.12
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{-(19.6)-\sqrt{1513.12}}{2*-4.8}=\frac{-19.6-\sqrt{1513.12}}{-9.6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19.6)+\sqrt{1513.12}}{2*-4.8}=\frac{-19.6+\sqrt{1513.12}}{-9.6}
$