D(t)=-t^2+6t

Solution for D(t)=-t^2+6t equation:

(D)=-D^2+6D

We move all terms to the left:

(D)-(-D^2+6D)=0

We get rid of parentheses

D^2-6D+D=0

We add all the numbers together, and all the variables

D^2-5D=0

a = 1; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·1·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*1}=\frac{0}{2}
=0
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*1}=\frac{10}{2}
=5
$