4.9t2+340t-2720=0

Solution for 4.9t2+340t-2720=0 equation:

4.9t^2+340t-2720=0

a = 4.9; b = 340; c = -2720;
Δ = b2-4ac
Δ = 3402-4·4.9·(-2720)
Δ = 168912
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:

$\sqrt{\Delta}=\sqrt{168912}=\sqrt{144*1173}=\sqrt{144}*\sqrt{1173}=12\sqrt{1173}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(340)-12\sqrt{1173}}{2*4.9}=\frac{-340-12\sqrt{1173}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(340)+12\sqrt{1173}}{2*4.9}=\frac{-340+12\sqrt{1173}}{9.8}
$