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

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

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

a = -16; b = 64; c = +16;
Δ = b2-4ac
Δ = 642-4·(-16)·16
Δ = 5120
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:

$\sqrt{\Delta}=\sqrt{5120}=\sqrt{1024*5}=\sqrt{1024}*\sqrt{5}=32\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(64)-32\sqrt{5}}{2*-16}=\frac{-64-32\sqrt{5}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(64)+32\sqrt{5}}{2*-16}=\frac{-64+32\sqrt{5}}{-32}
$