Show that if $a > 0$, then $ax^2 + bx + c ≥ 0$ for all values of $x$ if and only if $b^2 − 4ac ≤ 0$. So I'm fairly new to proofs and was wondering how to solve this problem. This is my train of thought.

I listed the terms as so: $ax^2 + bx + c ≥ 0 ≥ b^2 − 4ac$. 

From there I tried to complete the square and ended up with this:

$$x ≥ \sqrt{-c/a} ≥ \sqrt{(b^2/a) - 4ac - c/a}.$$

I'm not quite exactly sure how to show that if $a &gt; 0$, then $ax^2 + bx + c ≥ 0$ for all values of $x$ if and only if $b^2 − 4ac ≤ 0$ and would greatly enjoy any help.

Thanks. Forgive my formatting.
Note that $$ax^2+bx+c=a\cdot\left(x+\frac b{2a}\right)^2+c-\frac{b^2}{4a}=a\cdot\left(x+\frac b{2a}\right)^2-\frac{b^2-4ac}{4a}.$$
1.If the $b^2-4ac &lt; 0$ then equation has no real root so graph of the $ax^2+bx+c$ will always be above $X-axis$.
  
  2.If $b^2-4ac=0$ then equation will have only one real root.
  
  3.If $ax^2+bx+c$ is Non negative then it has at most one real root so $b^2-4ac$ is nonpositive


so $ax^2+bx+c$ is always non negative iff $b^2-4ac$ is non positive