4.9t^2+120t-2240=0

Solution for 4.9t^2+120t-2240=0 equation:

4.9t^2+120t-2240=0

a = 4.9; b = 120; c = -2240;
Δ = b2-4ac
Δ = 1202-4·4.9·(-2240)
Δ = 58304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{58304}=\sqrt{64*911}=\sqrt{64}*\sqrt{911}=8\sqrt{911}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(120)-8\sqrt{911}}{2*4.9}=\frac{-120-8\sqrt{911}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(120)+8\sqrt{911}}{2*4.9}=\frac{-120+8\sqrt{911}}{9.8}
$