D(t)=(20+30t)t

Solution for D(t)=(20+30t)t equation:

(D)=(20+30D)D

We move all terms to the left:

(D)-((20+30D)D)=0

We add all the numbers together, and all the variables

D-((30D+20)D)=0


We calculate terms in parentheses: -((30D+20)D), so:

(30D+20)D

We multiply parentheses

30D^2+20D

Back to the equation:

-(30D^2+20D)


We get rid of parentheses

-30D^2+D-20D=0

We add all the numbers together, and all the variables

-30D^2-19D=0

a = -30; b = -19; c = 0;
Δ = b2-4ac
Δ = -192-4·(-30)·0
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-19}{2*-30}=\frac{0}{-60}
=0
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+19}{2*-30}=\frac{38}{-60}
=-19/30
$