-16t^2+320+32=0

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

-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}
=-\frac{\sqrt{22}}{-1}
$
$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}
=\frac{\sqrt{22}}{-1}
$