1=16t^2+96t

Solution for 1=16t^2+96t equation:

1=16t^2+96t

We move all terms to the left:

1-(16t^2+96t)=0

We get rid of parentheses

-16t^2-96t+1=0

a = -16; b = -96; c = +1;
Δ = b2-4ac
Δ = -962-4·(-16)·1
Δ = 9280
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:

$\sqrt{\Delta}=\sqrt{9280}=\sqrt{64*145}=\sqrt{64}*\sqrt{145}=8\sqrt{145}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-96)-8\sqrt{145}}{2*-16}=\frac{96-8\sqrt{145}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-96)+8\sqrt{145}}{2*-16}=\frac{96+8\sqrt{145}}{-32}
$