16t^2-32t+8=0

Solution for 16t^2-32t+8=0 equation:

16t^2-32t+8=0

a = 16; b = -32; c = +8;
Δ = b2-4ac
Δ = -322-4·16·8
Δ = 512
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:

$\sqrt{\Delta}=\sqrt{512}=\sqrt{256*2}=\sqrt{256}*\sqrt{2}=16\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-16\sqrt{2}}{2*16}=\frac{32-16\sqrt{2}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+16\sqrt{2}}{2*16}=\frac{32+16\sqrt{2}}{32}
$