D=100t+4.9t^2

Solution for D=100t+4.9t^2 equation:

=100D+4.9D^2

We move all terms to the left:

-(100D+4.9D^2)=0

We get rid of parentheses

-4.9D^2-100D=0

a = -4.9; b = -100; c = 0;
Δ = b2-4ac
Δ = -1002-4·(-4.9)·0
Δ = 10000
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{10000}=100$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-100)-100}{2*-4.9}=\frac{0}{-9.8}
=0
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-100)+100}{2*-4.9}=\frac{200}{-9.8}
=-20+4/9.8
$