0=-16t^2+320+32

Solution for 0=-16t^2+320+32 equation:

0=-16t^2+320+32

We move all terms to the left:

0-(-16t^2+320+32)=0

We add all the numbers together, and all the variables

-(-16t^2+320+32)=0

We get rid of parentheses

16t^2-320-32=0

We add all the numbers together, and all the variables

16t^2-352=0

a = 16; b = 0; c = -352;
Δ = b2-4ac
Δ = 02-4·16·(-352)
Δ = 22528
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:

$\sqrt{\Delta}=\sqrt{22528}=\sqrt{1024*22}=\sqrt{1024}*\sqrt{22}=32\sqrt{22}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32\sqrt{22}}{2*16}=\frac{0-32\sqrt{22}}{32}
=-\frac{32\sqrt{22}}{32}
=-\sqrt{22}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32\sqrt{22}}{2*16}=\frac{0+32\sqrt{22}}{32}
=\frac{32\sqrt{22}}{32}
=\sqrt{22}
$