320=212+86t-16t^2

Solution for 320=212+86t-16t^2 equation:

320=212+86t-16t^2

We move all terms to the left:

320-(212+86t-16t^2)=0

We get rid of parentheses

16t^2-86t-212+320=0

We add all the numbers together, and all the variables

16t^2-86t+108=0

a = 16; b = -86; c = +108;
Δ = b2-4ac
Δ = -862-4·16·108
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-86)-22}{2*16}=\frac{64}{32}
=2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-86)+22}{2*16}=\frac{108}{32}
=3+3/8
$