-4.9t^2-8t+50=0

Solution for -4.9t^2-8t+50=0 equation:

-4.9t^2-8t+50=0

a = -4.9; b = -8; c = +50;
Δ = b2-4ac
Δ = -82-4·(-4.9)·50
Δ = 1044
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:

$\sqrt{\Delta}=\sqrt{1044}=\sqrt{36*29}=\sqrt{36}*\sqrt{29}=6\sqrt{29}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-6\sqrt{29}}{2*-4.9}=\frac{8-6\sqrt{29}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+6\sqrt{29}}{2*-4.9}=\frac{8+6\sqrt{29}}{-9.8}
$