D(t)=-49t^2+14.7t+49

Solution for D(t)=-49t^2+14.7t+49 equation:

(D)=-49D^2+14.7D+49

We move all terms to the left:

(D)-(-49D^2+14.7D+49)=0

We get rid of parentheses

49D^2-14.7D+D-49=0

We add all the numbers together, and all the variables

49D^2-13.7D-49=0

a = 49; b = -13.7; c = -49;
Δ = b2-4ac
Δ = -13.72-4·49·(-49)
Δ = 9791.69
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{-(-13.7)-\sqrt{9791.69}}{2*49}=\frac{13.7-\sqrt{9791.69}}{98}
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13.7)+\sqrt{9791.69}}{2*49}=\frac{13.7+\sqrt{9791.69}}{98}
$