0=-16t^2+48t+60

Solution for 0=-16t^2+48t+60 equation:

0=-16t^2+48t+60

We move all terms to the left:

0-(-16t^2+48t+60)=0

We add all the numbers together, and all the variables

-(-16t^2+48t+60)=0

We get rid of parentheses

16t^2-48t-60=0

a = 16; b = -48; c = -60;
Δ = b2-4ac
Δ = -482-4·16·(-60)
Δ = 6144
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:

$\sqrt{\Delta}=\sqrt{6144}=\sqrt{1024*6}=\sqrt{1024}*\sqrt{6}=32\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-32\sqrt{6}}{2*16}=\frac{48-32\sqrt{6}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+32\sqrt{6}}{2*16}=\frac{48+32\sqrt{6}}{32}
$