0=-16t^2+48t+80

Solution for 0=-16t^2+48t+80 equation:

0=-16t^2+48t+80

We move all terms to the left:

0-(-16t^2+48t+80)=0

We add all the numbers together, and all the variables

-(-16t^2+48t+80)=0

We get rid of parentheses

16t^2-48t-80=0

a = 16; b = -48; c = -80;
Δ = b2-4ac
Δ = -482-4·16·(-80)
Δ = 7424
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:

$\sqrt{\Delta}=\sqrt{7424}=\sqrt{256*29}=\sqrt{256}*\sqrt{29}=16\sqrt{29}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-16\sqrt{29}}{2*16}=\frac{48-16\sqrt{29}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+16\sqrt{29}}{2*16}=\frac{48+16\sqrt{29}}{32}
$