50=-14t+4.9t^2

Solution for 50=-14t+4.9t^2 equation:

50=-14t+4.9t^2

We move all terms to the left:

50-(-14t+4.9t^2)=0

We get rid of parentheses

-4.9t^2+14t+50=0

a = -4.9; b = 14; c = +50;
Δ = b2-4ac
Δ = 142-4·(-4.9)·50
Δ = 1176
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:

$\sqrt{\Delta}=\sqrt{1176}=\sqrt{196*6}=\sqrt{196}*\sqrt{6}=14\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-14\sqrt{6}}{2*-4.9}=\frac{-14-14\sqrt{6}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+14\sqrt{6}}{2*-4.9}=\frac{-14+14\sqrt{6}}{-9.8}
$