D(5)=0.8t^2

Solution for D(5)=0.8t^2 equation:

(5)=0.8D^2

We move all terms to the left:

(5)-(0.8D^2)=0

We get rid of parentheses

-0.8D^2+5=0

a = -0.8; b = 0; c = +5;
Δ = b2-4ac
Δ = 02-4·(-0.8)·5
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4}{2*-0.8}=\frac{-4}{-1.6}
=2+0.8/1.6
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4}{2*-0.8}=\frac{4}{-1.6}
=-2+0.8/1.6
$