0=-16t^2+12t+120

Solution for 0=-16t^2+12t+120 equation:

0=-16t^2+12t+120

We move all terms to the left:

0-(-16t^2+12t+120)=0

We add all the numbers together, and all the variables

-(-16t^2+12t+120)=0

We get rid of parentheses

16t^2-12t-120=0

a = 16; b = -12; c = -120;
Δ = b2-4ac
Δ = -122-4·16·(-120)
Δ = 7824
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:

$\sqrt{\Delta}=\sqrt{7824}=\sqrt{16*489}=\sqrt{16}*\sqrt{489}=4\sqrt{489}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{489}}{2*16}=\frac{12-4\sqrt{489}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{489}}{2*16}=\frac{12+4\sqrt{489}}{32}
$