-4.9t^2+120t-100=0

Solution for -4.9t^2+120t-100=0 equation:

-4.9t^2+120t-100=0

a = -4.9; b = 120; c = -100;
Δ = b2-4ac
Δ = 1202-4·(-4.9)·(-100)
Δ = 12440
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:

$\sqrt{\Delta}=\sqrt{12440}=\sqrt{4*3110}=\sqrt{4}*\sqrt{3110}=2\sqrt{3110}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(120)-2\sqrt{3110}}{2*-4.9}=\frac{-120-2\sqrt{3110}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(120)+2\sqrt{3110}}{2*-4.9}=\frac{-120+2\sqrt{3110}}{-9.8}
$