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.

Note that $a$ and $b$ only appear in the expression $a+b$. So let us simplify our lives and denote $a+b =:d $, so we need to show that the following has real roots:
$$ (d-c)x^2 + 2dx + (c+d) = 0.$$
There is a useful object called the discriminant. For an equation $Ax^2 + Bx + C$ it is given as: $\Delta = B^2 - 4 AC$. The roots of a polynomial are real iff $\Delta \geq 0$. For our polynomial, we have:
$$\Delta = 4d^2 - 4(d+c)(d-c) = 4 c^2 \geq 0.  $$
Because $\Delta \geq 0$, the roots are real, and you are done.

By the way, the roots of  $Ax^2 + Bx + C$ are $\frac{-B \pm \Delta}{2A}$, which is very often very useful.
