m^2-8m+32=28

Solution for m^2-8m+32=28 equation:

m^2-8m+32=28

We move all terms to the left:

m^2-8m+32-(28)=0

We add all the numbers together, and all the variables

m^2-8m+4=0

a = 1; b = -8; c = +4;
Δ = b2-4ac
Δ = -82-4·1·4
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4\sqrt{3}}{2*1}=\frac{8-4\sqrt{3}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4\sqrt{3}}{2*1}=\frac{8+4\sqrt{3}}{2}
$