If you have a polynomial #f(x)# of degree #n# and you can find all its solution #x_1,..,x_n#, then you have that
#f(x)=(x-x_1)* ... *(x-x_n)#
In your case, #n=2#, so if we find two solutions #x_1# and #x_2#, we can write
#x^2-8x+6=(x-x_1)(x-x_2)#
To find the solution, we can use the standard formula
#x_{1,2} = \frac{-b\pm\sqrt(b^2-4ac)}{2a}#
In your case, #a=1#, #b=-8# and #c=6#. So, the formula becomes
#x_{1,2} = \frac{8\pm\sqrt(64-4*1*6)}{2}
= \frac{8\pm\sqrt(64-24)}{2}
=\frac{8\pm\sqrt(40)}{2}#
Since #sqrt(40)=sqrt(4*10)=sqrt(4)*sqrt(10)=2sqrt(10)#, the formula becomes
#\frac{8\pm2\sqrt(10)}{2} = 4\pmsqrt(10) #
So, the solutions are #x_1 = 4+sqrt(10)# and #x_2 = 4-sqrt(10)#, and thus we have the factorization
#x^2-8x+6=(x-4-sqrt(10))(x-4+sqrt(10))#
