-4.9t^2-3t+200=0

Solution for -4.9t^2-3t+200=0 equation:

-4.9t^2-3t+200=0

a = -4.9; b = -3; c = +200;
Δ = b2-4ac
Δ = -32-4·(-4.9)·200
Δ = 3929
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:

$\sqrt{\Delta}=\sqrt{3929}=\sqrt{1*3929}=\sqrt{1}*\sqrt{3929}=1\sqrt{3929}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-1\sqrt{3929}}{2*-4.9}=\frac{3-1\sqrt{3929}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+1\sqrt{3929}}{2*-4.9}=\frac{3+1\sqrt{3929}}{-9.8}
$