14t^2-8t+1=0

Solution for 14t^2-8t+1=0 equation:

14t^2-8t+1=0

a = 14; b = -8; c = +1;
Δ = b2-4ac
Δ = -82-4·14·1
Δ = 8
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:

$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{2}}{2*14}=\frac{8-2\sqrt{2}}{28}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{2}}{2*14}=\frac{8+2\sqrt{2}}{28}
$