-16t^2-8t+64=0

Solution for -16t^2-8t+64=0 equation:

-16t^2-8t+64=0

a = -16; b = -8; c = +64;
Δ = b2-4ac
Δ = -82-4·(-16)·64
Δ = 4160
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:

$\sqrt{\Delta}=\sqrt{4160}=\sqrt{64*65}=\sqrt{64}*\sqrt{65}=8\sqrt{65}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8\sqrt{65}}{2*-16}=\frac{8-8\sqrt{65}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8\sqrt{65}}{2*-16}=\frac{8+8\sqrt{65}}{-32}
$