D(t)=-2t^2+9t+18

Solution for D(t)=-2t^2+9t+18 equation:

(D)=-2D^2+9D+18

We move all terms to the left:

(D)-(-2D^2+9D+18)=0

We get rid of parentheses

2D^2-9D+D-18=0

We add all the numbers together, and all the variables

2D^2-8D-18=0

a = 2; b = -8; c = -18;
Δ = b2-4ac
Δ = -82-4·2·(-18)
Δ = 208
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4\sqrt{13}}{2*2}=\frac{8-4\sqrt{13}}{4}
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4\sqrt{13}}{2*2}=\frac{8+4\sqrt{13}}{4}
$