Prove that if $ a,b,c > 0$, then if $ax^{2} + bx + c = 0$ has real solutions, both solutions are negative. I want to prove this by contradiction.

Proof: If we let $a,b,c = 1$, we have $x^{2} + x + 1 = 0$

or $x^{2} + x = -1$. If $x$ is positive the sum of two positive terms can not be $< 0$

hence $x$ must be negative. But no matter the $x$ we pick there will be no solution to $x^{2} + x = -1$.

So $a,b,c$ must also be $ < 0$. If $a,b,c < 0$ we have $-ax^{2} -bx -c = 0$

or $-ax^{2} -bx = c$. If $x$ is positive then resulting equation is $negative - negative = positive$, which is not possible. So $x$ must be negative.

Have I shown by contradiction, both solutions are negative? How can I improve my answer?