D=3.5+80+-16t^2

Solution for D=3.5+80+-16t^2 equation:

=3.5+80+-16D^2

We move all terms to the left:

-(3.5+80+-16D^2)=0

We use the square of the difference formula

-(3.5+80-16D^2)=0

We get rid of parentheses

16D^2-3.5-80=0

We add all the numbers together, and all the variables

16D^2-83.5=0

a = 16; b = 0; c = -83.5;
Δ = b2-4ac
Δ = 02-4·16·(-83.5)
Δ = 5344
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:

$\sqrt{\Delta}=\sqrt{5344}=\sqrt{16*334}=\sqrt{16}*\sqrt{334}=4\sqrt{334}$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{334}}{2*16}=\frac{0-4\sqrt{334}}{32}
=-\frac{4\sqrt{334}}{32}
=-\frac{\sqrt{334}}{8}
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{334}}{2*16}=\frac{0+4\sqrt{334}}{32}
=\frac{4\sqrt{334}}{32}
=\frac{\sqrt{334}}{8}
$