D=-16t^2-4t+412

Solution for D=-16t^2-4t+412 equation:

=-16D^2-4D+412

We move all terms to the left:

-(-16D^2-4D+412)=0

We get rid of parentheses

16D^2+4D-412=0

a = 16; b = 4; c = -412;
Δ = b2-4ac
Δ = 42-4·16·(-412)
Δ = 26384
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:

$\sqrt{\Delta}=\sqrt{26384}=\sqrt{16*1649}=\sqrt{16}*\sqrt{1649}=4\sqrt{1649}$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{1649}}{2*16}=\frac{-4-4\sqrt{1649}}{32}
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{1649}}{2*16}=\frac{-4+4\sqrt{1649}}{32}
$