4.9t^2+0.5t-200=0

Solution for 4.9t^2+0.5t-200=0 equation:

4.9t^2+0.5t-200=0

a = 4.9; b = 0.5; c = -200;
Δ = b2-4ac
Δ = 0.52-4·4.9·(-200)
Δ = 3920.25
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{-(0.5)-\sqrt{3920.25}}{2*4.9}=\frac{-0.5-\sqrt{3920.25}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.5)+\sqrt{3920.25}}{2*4.9}=\frac{-0.5+\sqrt{3920.25}}{9.8}
$