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

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

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

We add all the numbers together, and all the variables

-16t^2+64t-62=0

a = -16; b = 64; c = -62;
Δ = b2-4ac
Δ = 642-4·(-16)·(-62)
Δ = 128
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:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(64)-8\sqrt{2}}{2*-16}=\frac{-64-8\sqrt{2}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(64)+8\sqrt{2}}{2*-16}=\frac{-64+8\sqrt{2}}{-32}
$