16=-16t^2+32+4

Solution for 16=-16t^2+32+4 equation:

16=-16t^2+32+4

We move all terms to the left:

16-(-16t^2+32+4)=0

We get rid of parentheses

16t^2-32-4+16=0

We add all the numbers together, and all the variables

16t^2-20=0

a = 16; b = 0; c = -20;
Δ = b2-4ac
Δ = 02-4·16·(-20)
Δ = 1280
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1280}=\sqrt{256*5}=\sqrt{256}*\sqrt{5}=16\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{5}}{2*16}=\frac{0-16\sqrt{5}}{32}
=-\frac{16\sqrt{5}}{32}
=-\frac{\sqrt{5}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{5}}{2*16}=\frac{0+16\sqrt{5}}{32}
=\frac{16\sqrt{5}}{32}
=\frac{\sqrt{5}}{2}
$