160=16t-0.2t^2

Solution for 160=16t-0.2t^2 equation:

160=16t-0.2t^2

We move all terms to the left:

160-(16t-0.2t^2)=0

We get rid of parentheses

0.2t^2-16t+160=0

a = 0.2; b = -16; c = +160;
Δ = b2-4ac
Δ = -162-4·0.2·160
Δ = 128
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:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-8\sqrt{2}}{2*0.2}=\frac{16-8\sqrt{2}}{0.4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+8\sqrt{2}}{2*0.2}=\frac{16+8\sqrt{2}}{0.4}
$