320=16t2+4t

Solution for 320=16t2+4t equation:

320=16t^2+4t

We move all terms to the left:

320-(16t^2+4t)=0

We get rid of parentheses

-16t^2-4t+320=0

a = -16; b = -4; c = +320;
Δ = b2-4ac
Δ = -42-4·(-16)·320
Δ = 20496
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:

$\sqrt{\Delta}=\sqrt{20496}=\sqrt{16*1281}=\sqrt{16}*\sqrt{1281}=4\sqrt{1281}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{1281}}{2*-16}=\frac{4-4\sqrt{1281}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{1281}}{2*-16}=\frac{4+4\sqrt{1281}}{-32}
$