-120-8t+4.9t^2=0

Solution for -120-8t+4.9t^2=0 equation:

-120-8t+4.9t^2=0

a = 4.9; b = -8; c = -120;
Δ = b2-4ac
Δ = -82-4·4.9·(-120)
Δ = 2416
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:

$\sqrt{\Delta}=\sqrt{2416}=\sqrt{16*151}=\sqrt{16}*\sqrt{151}=4\sqrt{151}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4\sqrt{151}}{2*4.9}=\frac{8-4\sqrt{151}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4\sqrt{151}}{2*4.9}=\frac{8+4\sqrt{151}}{9.8}
$