4.9t^2+9t-500=0

Solution for 4.9t^2+9t-500=0 equation:

4.9t^2+9t-500=0

a = 4.9; b = 9; c = -500;
Δ = b2-4ac
Δ = 92-4·4.9·(-500)
Δ = 9881
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{-(9)-\sqrt{9881}}{2*4.9}=\frac{-9-\sqrt{9881}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{9881}}{2*4.9}=\frac{-9+\sqrt{9881}}{9.8}
$