D(t)=0.33t^2

Solution for D(t)=0.33t^2 equation:

(D)=0.33D^2

We move all terms to the left:

(D)-(0.33D^2)=0

We get rid of parentheses

-0.33D^2+D=0

a = -0.33; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-0.33)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*-0.33}=\frac{-2}{-0.66}
=3+0.02/0.66
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-0.33}=\frac{0}{-0.66}
=0
$