50+4t-4.9t^2=0

Solution for 50+4t-4.9t^2=0 equation:

50+4t-4.9t^2=0

a = -4.9; b = 4; c = +50;
Δ = b2-4ac
Δ = 42-4·(-4.9)·50
Δ = 996
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:

$\sqrt{\Delta}=\sqrt{996}=\sqrt{4*249}=\sqrt{4}*\sqrt{249}=2\sqrt{249}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{249}}{2*-4.9}=\frac{-4-2\sqrt{249}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{249}}{2*-4.9}=\frac{-4+2\sqrt{249}}{-9.8}
$