600t-4.9t^2+20=0

Solution for 600t-4.9t^2+20=0 equation:

600t-4.9t^2+20=0

a = -4.9; b = 600; c = +20;
Δ = b2-4ac
Δ = 6002-4·(-4.9)·20
Δ = 360392
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:

$\sqrt{\Delta}=\sqrt{360392}=\sqrt{4*90098}=\sqrt{4}*\sqrt{90098}=2\sqrt{90098}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(600)-2\sqrt{90098}}{2*-4.9}=\frac{-600-2\sqrt{90098}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(600)+2\sqrt{90098}}{2*-4.9}=\frac{-600+2\sqrt{90098}}{-9.8}
$