D(3)=4.9t^2

Solution for D(3)=4.9t^2 equation:

(3)=4.9D^2

We move all terms to the left:

(3)-(4.9D^2)=0

We get rid of parentheses

-4.9D^2+3=0

a = -4.9; b = 0; c = +3;
Δ = b2-4ac
Δ = 02-4·(-4.9)·3
Δ = 58.8
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-\sqrt{58.8}}{2*-4.9}=\frac{0-\sqrt{58.8}}{-9.8}
=-\frac{\sqrt{}}{-9.8}
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{58.8}}{2*-4.9}=\frac{0+\sqrt{58.8}}{-9.8}
=\frac{\sqrt{}}{-9.8}
$