0=-16t2+48+64

Solution for 0=-16t2+48+64 equation:

0=-16t^2+48+64

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

16t^2-48-64=0

We add all the numbers together, and all the variables

16t^2-112=0

a = 16; b = 0; c = -112;
Δ = b2-4ac
Δ = 02-4·16·(-112)
Δ = 7168
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:

$\sqrt{\Delta}=\sqrt{7168}=\sqrt{1024*7}=\sqrt{1024}*\sqrt{7}=32\sqrt{7}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32\sqrt{7}}{2*16}=\frac{0-32\sqrt{7}}{32}
=-\frac{32\sqrt{7}}{32}
=-\sqrt{7}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32\sqrt{7}}{2*16}=\frac{0+32\sqrt{7}}{32}
=\frac{32\sqrt{7}}{32}
=\sqrt{7}
$