4.9t^2+340t-1360=0

Solution for 4.9t^2+340t-1360=0 equation:

4.9t^2+340t-1360=0

a = 4.9; b = 340; c = -1360;
Δ = b2-4ac
Δ = 3402-4·4.9·(-1360)
Δ = 142256
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:

$\sqrt{\Delta}=\sqrt{142256}=\sqrt{16*8891}=\sqrt{16}*\sqrt{8891}=4\sqrt{8891}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(340)-4\sqrt{8891}}{2*4.9}=\frac{-340-4\sqrt{8891}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(340)+4\sqrt{8891}}{2*4.9}=\frac{-340+4\sqrt{8891}}{9.8}
$