0=-16t^2+60t+120

Solution for 0=-16t^2+60t+120 equation:

0=-16t^2+60t+120

We move all terms to the left:

0-(-16t^2+60t+120)=0

We add all the numbers together, and all the variables

-(-16t^2+60t+120)=0

We get rid of parentheses

16t^2-60t-120=0

a = 16; b = -60; c = -120;
Δ = b2-4ac
Δ = -602-4·16·(-120)
Δ = 11280
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:

$\sqrt{\Delta}=\sqrt{11280}=\sqrt{16*705}=\sqrt{16}*\sqrt{705}=4\sqrt{705}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-60)-4\sqrt{705}}{2*16}=\frac{60-4\sqrt{705}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-60)+4\sqrt{705}}{2*16}=\frac{60+4\sqrt{705}}{32}
$