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}
$