Prove that roots are real I am stuck with this equation, I need to prove the roots are real when $a, b, c \in R$
The equation is $$(a+b-c)x^2+ 2(a+b)x + (a+b+c) = 0$$
If someone could tell me the right way to go about this, so I can attempt it.
Thank you
EDIT: I have made an error in the question.  I have now corrected it.

Set $a+b-c=A$ and $a+b+c=B$; then $2(a+b)=A+B$ and your equation is
$$
Ax^2+(A+B)x+B=0
$$
that can be written
$$
Ax^2+Ax+Bx+B=0
$$
or
$$
Ax(x+1)+B(x+1)=0
$$
and finally
$$
(Ax+B)(x+1)=0.
$$
If $A=0$, then the equation is $B(x+1)=0$. If also $B=0$, the equation is an identity; otherwise it has the only solution $-1$. If $A\ne0$, the solutions are $-1$ and $-B/A$.
