0=-16t^2-48t+64

Solution for 0=-16t^2-48t+64 equation:

0=-16t^2-48t+64

We move all terms to the left:

0-(-16t^2-48t+64)=0

We add all the numbers together, and all the variables

-(-16t^2-48t+64)=0

We get rid of parentheses

16t^2+48t-64=0

a = 16; b = 48; c = -64;
Δ = b2-4ac
Δ = 482-4·16·(-64)
Δ = 6400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6400}=80$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-80}{2*16}=\frac{-128}{32}
=-4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+80}{2*16}=\frac{32}{32}
=1
$