-4.9t^2+19t+50=0

Solution for -4.9t^2+19t+50=0 equation:

-4.9t^2+19t+50=0

a = -4.9; b = 19; c = +50;
Δ = b2-4ac
Δ = 192-4·(-4.9)·50
Δ = 1341
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:

$\sqrt{\Delta}=\sqrt{1341}=\sqrt{9*149}=\sqrt{9}*\sqrt{149}=3\sqrt{149}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-3\sqrt{149}}{2*-4.9}=\frac{-19-3\sqrt{149}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+3\sqrt{149}}{2*-4.9}=\frac{-19+3\sqrt{149}}{-9.8}
$