Proving that the quadratic equation $(q-5)x^2 +5x -q=0$ has real roots for any value of $q$

Question

Prove that the quadratic equation $(q-5)x^2 +5x -q=0$ has real roots for any value of $q$.

So I have already tried using the discriminant but just wanted to see if my answer is right or not. Appreciate every feedback.

Answer 1

$$Discriminant = {\sqrt {b^2-4ac}}$$ $$ = {\sqrt {5^2-4(-q)(q-5)}}$$ $$ = {\sqrt {25+4(q)(q-5)}}$$ $$ = {\sqrt {4q^2-20q+25}}$$ $$ = {\sqrt {(2q-5)^2}}$$

As ${(2q-5)^2}$ ${\ge}\,0$, real roots always exist.

Answer 2

As a hint, and supplementing the discriminant solution offered previously: $$(q-5)x^2 +5x -q \:=\: (x-1)\big((q-5)x+q\big)$$