16t^2+8t-1=0

Solution for 16t^2+8t-1=0 equation:

16t^2+8t-1=0

a = 16; b = 8; c = -1;
Δ = b2-4ac
Δ = 82-4·16·(-1)
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{-(8)-8\sqrt{2}}{2*16}=\frac{-8-8\sqrt{2}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{2}}{2*16}=\frac{-8+8\sqrt{2}}{32}
$