4.9t^2-40t-98=0

Solution for 4.9t^2-40t-98=0 equation:

4.9t^2-40t-98=0

a = 4.9; b = -40; c = -98;
Δ = b2-4ac
Δ = -402-4·4.9·(-98)
Δ = 3520.8
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{-(-40)-\sqrt{3520.8}}{2*4.9}=\frac{40-\sqrt{3520.8}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+\sqrt{3520.8}}{2*4.9}=\frac{40+\sqrt{3520.8}}{9.8}
$