m^2-26m+41=0

Solution for m^2-26m+41=0 equation:

m^2-26m+41=0

a = 1; b = -26; c = +41;
Δ = b2-4ac
Δ = -262-4·1·41
Δ = 512
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:

$\sqrt{\Delta}=\sqrt{512}=\sqrt{256*2}=\sqrt{256}*\sqrt{2}=16\sqrt{2}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-16\sqrt{2}}{2*1}=\frac{26-16\sqrt{2}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+16\sqrt{2}}{2*1}=\frac{26+16\sqrt{2}}{2}
$