16t^2-64t-8=0

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

16t^2-64t-8=0

a = 16; b = -64; c = -8;
Δ = b2-4ac
Δ = -642-4·16·(-8)
Δ = 4608
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:

$\sqrt{\Delta}=\sqrt{4608}=\sqrt{2304*2}=\sqrt{2304}*\sqrt{2}=48\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-64)-48\sqrt{2}}{2*16}=\frac{64-48\sqrt{2}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-64)+48\sqrt{2}}{2*16}=\frac{64+48\sqrt{2}}{32}
$