-4.9t^2+14t+500=0

Solution for -4.9t^2+14t+500=0 equation:

-4.9t^2+14t+500=0

a = -4.9; b = 14; c = +500;
Δ = b2-4ac
Δ = 142-4·(-4.9)·500
Δ = 9996
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:

$\sqrt{\Delta}=\sqrt{9996}=\sqrt{196*51}=\sqrt{196}*\sqrt{51}=14\sqrt{51}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-14\sqrt{51}}{2*-4.9}=\frac{-14-14\sqrt{51}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+14\sqrt{51}}{2*-4.9}=\frac{-14+14\sqrt{51}}{-9.8}
$