Expanding out the equation, we get
[tex]x^2 - ax - bx+ab+x^2 -bx-cx+bc+x^2 -ax-cx+ac=0 \\ \\ 3x^2 -x(2a+2b+2c)+(ab+bc+ac)=0[/tex]
Considering the discriminant,
[tex](2a+2b+2c)^2 - 4(3)(ab+bc+ac) \\ \\ =4a^2 + 4b^2 + 4c^2 + 8(ab+bc+ac)-12(ab+bc+ac) \\ \\ =4(a^2 + b^2 + c^2 - ab - bc - ac) \\ \\ =2((a-b)^2 + (b-c)^2 + (c-a)^2)[/tex]
This is always non-negative, meaning the roots are real.
The roots are equal if and only if the discriminant is 0, which is when a=b=c.
