1=-16t^2+8

Solution for 1=-16t^2+8 equation:

1=-16t^2+8

We move all terms to the left:

1-(-16t^2+8)=0

We get rid of parentheses

16t^2-8+1=0

We add all the numbers together, and all the variables

16t^2-7=0

a = 16; b = 0; c = -7;
Δ = b2-4ac
Δ = 02-4·16·(-7)
Δ = 448
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:

$\sqrt{\Delta}=\sqrt{448}=\sqrt{64*7}=\sqrt{64}*\sqrt{7}=8\sqrt{7}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{7}}{2*16}=\frac{0-8\sqrt{7}}{32}
=-\frac{8\sqrt{7}}{32}
=-\frac{\sqrt{7}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{7}}{2*16}=\frac{0+8\sqrt{7}}{32}
=\frac{8\sqrt{7}}{32}
=\frac{\sqrt{7}}{4}
$