4t^2-20t+1=0

Solution for 4t^2-20t+1=0 equation:

4t^2-20t+1=0

a = 4; b = -20; c = +1;
Δ = b2-4ac
Δ = -202-4·4·1
Δ = 384
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:

$\sqrt{\Delta}=\sqrt{384}=\sqrt{64*6}=\sqrt{64}*\sqrt{6}=8\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-8\sqrt{6}}{2*4}=\frac{20-8\sqrt{6}}{8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+8\sqrt{6}}{2*4}=\frac{20+8\sqrt{6}}{8}
$