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

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

(D)=16D-2D^2

We move all terms to the left:

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

We get rid of parentheses

2D^2-16D+D=0

We add all the numbers together, and all the variables

2D^2-15D=0

a = 2; b = -15; c = 0;
Δ = b2-4ac
Δ = -152-4·2·0
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-15}{2*2}=\frac{0}{4}
=0
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+15}{2*2}=\frac{30}{4}
=7+1/2
$