First divide through by 2 to get:
#x^2 - 6x + 11/2 = 0#
Now #(x - 3)^2# = #x^2 -6x + 9#
So we can write
#0 = x^2 - 6x + 11/2#
#= x^2 -6x + 9 - 9 + 11/2#
#= (x-3)^2 - 9 + 11/2#
#= (x - 3)^2 - 18/2 + 11/2#
#= (x - 3)^2 - 7/2#
Adding #7/2# to both sides we get
#(x - 3)^2 = 7/2#
So #(x - 3) = +- sqrt(7/2)#
Add 3 to both sides to get
#x = 3 +- sqrt(7/2) = 3 +- sqrt(7)/sqrt(2)#
If you prefer, #sqrt(7/2) = sqrt(14/4) = sqrt(14)/2#, so
#x = 3 +- sqrt(14)/2#
In general,
#ax^2 + bx + c = a(x + b/(2a))^2 + (c - b^2/(4a))#
Hence #ax^2 + bx + c = 0# has solutions
#x = (-b +-sqrt(b^2 - 4ac))/(2a)#
