80=16t^2+32

Solution for 80=16t^2+32 equation:

80=16t^2+32

We move all terms to the left:

80-(16t^2+32)=0

We get rid of parentheses

-16t^2-32+80=0

We add all the numbers together, and all the variables

-16t^2+48=0

a = -16; b = 0; c = +48;
Δ = b2-4ac
Δ = 02-4·(-16)·48
Δ = 3072
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:

$\sqrt{\Delta}=\sqrt{3072}=\sqrt{1024*3}=\sqrt{1024}*\sqrt{3}=32\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32\sqrt{3}}{2*-16}=\frac{0-32\sqrt{3}}{-32}
=-\frac{32\sqrt{3}}{-32}
=-\frac{\sqrt{3}}{-1}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32\sqrt{3}}{2*-16}=\frac{0+32\sqrt{3}}{-32}
=\frac{32\sqrt{3}}{-32}
=\frac{\sqrt{3}}{-1}
$