16u^2-12u-1=0

Solution for 16u^2-12u-1=0 equation:

16u^2-12u-1=0

a = 16; b = -12; c = -1;
Δ = b2-4ac
Δ = -122-4·16·(-1)
Δ = 208
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{13}}{2*16}=\frac{12-4\sqrt{13}}{32}
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{13}}{2*16}=\frac{12+4\sqrt{13}}{32}
$