40=16t-0.8t^2

Solution for 40=16t-0.8t^2 equation:

40=16t-0.8t^2

We move all terms to the left:

40-(16t-0.8t^2)=0

We get rid of parentheses

0.8t^2-16t+40=0

a = 0.8; b = -16; c = +40;
Δ = b2-4ac
Δ = -162-4·0.8·40
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*0.8}=\frac{16-8\sqrt{2}}{1.6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+8\sqrt{2}}{2*0.8}=\frac{16+8\sqrt{2}}{1.6}
$