0=-16t^2+128t+320

Solution for 0=-16t^2+128t+320 equation:

0=-16t^2+128t+320

We move all terms to the left:

0-(-16t^2+128t+320)=0

We add all the numbers together, and all the variables

-(-16t^2+128t+320)=0

We get rid of parentheses

16t^2-128t-320=0

a = 16; b = -128; c = -320;
Δ = b2-4ac
Δ = -1282-4·16·(-320)
Δ = 36864
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36864}=192$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-128)-192}{2*16}=\frac{-64}{32}
=-2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-128)+192}{2*16}=\frac{320}{32}
=10
$