D=-16t^2+435

Solution for D=-16t^2+435 equation:

=-16D^2+435

We move all terms to the left:

-(-16D^2+435)=0

We get rid of parentheses

16D^2-435=0

a = 16; b = 0; c = -435;
Δ = b2-4ac
Δ = 02-4·16·(-435)
Δ = 27840
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:

$\sqrt{\Delta}=\sqrt{27840}=\sqrt{64*435}=\sqrt{64}*\sqrt{435}=8\sqrt{435}$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{435}}{2*16}=\frac{0-8\sqrt{435}}{32}
=-\frac{8\sqrt{435}}{32}
=-\frac{\sqrt{435}}{4}
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{435}}{2*16}=\frac{0+8\sqrt{435}}{32}
=\frac{8\sqrt{435}}{32}
=\frac{\sqrt{435}}{4}
$