4.9t^2+2t-100=0

Solution for 4.9t^2+2t-100=0 equation:

4.9t^2+2t-100=0

a = 4.9; b = 2; c = -100;
Δ = b2-4ac
Δ = 22-4·4.9·(-100)
Δ = 1964
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:

$\sqrt{\Delta}=\sqrt{1964}=\sqrt{4*491}=\sqrt{4}*\sqrt{491}=2\sqrt{491}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{491}}{2*4.9}=\frac{-2-2\sqrt{491}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{491}}{2*4.9}=\frac{-2+2\sqrt{491}}{9.8}
$