-4.9t^2-16t+50=0

Solution for -4.9t^2-16t+50=0 equation:

-4.9t^2-16t+50=0

a = -4.9; b = -16; c = +50;
Δ = b2-4ac
Δ = -162-4·(-4.9)·50
Δ = 1236
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:

$\sqrt{\Delta}=\sqrt{1236}=\sqrt{4*309}=\sqrt{4}*\sqrt{309}=2\sqrt{309}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2\sqrt{309}}{2*-4.9}=\frac{16-2\sqrt{309}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2\sqrt{309}}{2*-4.9}=\frac{16+2\sqrt{309}}{-9.8}
$