D=-16t2+8t

Solution for D=-16t2+8t equation:

=-16D^2+8D

We move all terms to the left:

-(-16D^2+8D)=0

We get rid of parentheses

16D^2-8D=0

a = 16; b = -8; c = 0;
Δ = b2-4ac
Δ = -82-4·16·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8}{2*16}=\frac{0}{32}
=0
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8}{2*16}=\frac{16}{32}
=1/2
$