160=32t-0.8t^2

Solution for 160=32t-0.8t^2 equation:

160=32t-0.8t^2

We move all terms to the left:

160-(32t-0.8t^2)=0

We get rid of parentheses

0.8t^2-32t+160=0

a = 0.8; b = -32; c = +160;
Δ = b2-4ac
Δ = -322-4·0.8·160
Δ = 512
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:

$\sqrt{\Delta}=\sqrt{512}=\sqrt{256*2}=\sqrt{256}*\sqrt{2}=16\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-16\sqrt{2}}{2*0.8}=\frac{32-16\sqrt{2}}{1.6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+16\sqrt{2}}{2*0.8}=\frac{32+16\sqrt{2}}{1.6}
$