16t^2-12t+1=0

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

16t^2-12t+1=0

a = 16; b = -12; c = +1;
Δ = b2-4ac
Δ = -122-4·16·1
Δ = 80
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:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{5}}{2*16}=\frac{12-4\sqrt{5}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{5}}{2*16}=\frac{12+4\sqrt{5}}{32}
$