20x^2-16x+1=0

Solution for 20x^2-16x+1=0 equation:

20x^2-16x+1=0

a = 20; b = -16; c = +1;
Δ = b2-4ac
Δ = -162-4·20·1
Δ = 176
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{176}=\sqrt{16*11}=\sqrt{16}*\sqrt{11}=4\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4\sqrt{11}}{2*20}=\frac{16-4\sqrt{11}}{40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4\sqrt{11}}{2*20}=\frac{16+4\sqrt{11}}{40}
$