(m^2-4m+1)=0

Solution for (m^2-4m+1)=0 equation:

(m^2-4m+1)=0

We get rid of parentheses

m^2-4m+1=0

a = 1; b = -4; c = +1;
Δ = b2-4ac
Δ = -42-4·1·1
Δ = 12
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{3}}{2*1}=\frac{4-2\sqrt{3}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{3}}{2*1}=\frac{4+2\sqrt{3}}{2}
$