D(t)=-16t^2+400t

Solution for D(t)=-16t^2+400t equation:

(D)=-16D^2+400D

We move all terms to the left:

(D)-(-16D^2+400D)=0

We get rid of parentheses

16D^2-400D+D=0

We add all the numbers together, and all the variables

16D^2-399D=0

a = 16; b = -399; c = 0;
Δ = b2-4ac
Δ = -3992-4·16·0
Δ = 159201
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{159201}=399$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-399)-399}{2*16}=\frac{0}{32}
=0
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-399)+399}{2*16}=\frac{798}{32}
=24+15/16
$