m^2-16m+41=10

Solution for m^2-16m+41=10 equation:

m^2-16m+41=10

We move all terms to the left:

m^2-16m+41-(10)=0

We add all the numbers together, and all the variables

m^2-16m+31=0

a = 1; b = -16; c = +31;
Δ = b2-4ac
Δ = -162-4·1·31
Δ = 132
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:

$\sqrt{\Delta}=\sqrt{132}=\sqrt{4*33}=\sqrt{4}*\sqrt{33}=2\sqrt{33}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2\sqrt{33}}{2*1}=\frac{16-2\sqrt{33}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2\sqrt{33}}{2*1}=\frac{16+2\sqrt{33}}{2}
$