-0.8t^2+16t-40=0

Solution for -0.8t^2+16t-40=0 equation:

-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}
$