-4.9t^2+14t+98=0

Solution for -4.9t^2+14t+98=0 equation:

-4.9t^2+14t+98=0

a = -4.9; b = 14; c = +98;
Δ = b2-4ac
Δ = 142-4·(-4.9)·98
Δ = 2116.8
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{-(14)-\sqrt{2116.8}}{2*-4.9}=\frac{-14-\sqrt{2116.8}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+\sqrt{2116.8}}{2*-4.9}=\frac{-14+\sqrt{2116.8}}{-9.8}
$