-4.9t^2+40t+50=0

Solution for -4.9t^2+40t+50=0 equation:

-4.9t^2+40t+50=0

a = -4.9; b = 40; c = +50;
Δ = b2-4ac
Δ = 402-4·(-4.9)·50
Δ = 2580
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:

$\sqrt{\Delta}=\sqrt{2580}=\sqrt{4*645}=\sqrt{4}*\sqrt{645}=2\sqrt{645}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-2\sqrt{645}}{2*-4.9}=\frac{-40-2\sqrt{645}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+2\sqrt{645}}{2*-4.9}=\frac{-40+2\sqrt{645}}{-9.8}
$