If I let $y=ax^2+bx+c, (a\neq 0)$ then extremum of $y$ is attained at $x=-\frac{b}{2a}$.
Then $\large\frac{\mathrm {d^2}y}{\mathrm {d}x^2}\big|_{(x=-\frac{b}{2a})}=2a$ which is positive or negative accordingly as $a>0$ or $a<0$ respectively. For $a>0$ local minimum of $y$ is obtained and the minimum value is $y(\frac{1}{2})=\frac{4ac-b^2}{4a}$. For $x=-\frac{b}{2a}$ the inequality $y\geq 0$ must also hold, whence $\frac{4ac-b^2}{4a}\geq 0$.
For $a>0$, $b^2\leq 4ac$ and for $a<0$, $b^2\geq 4ac$ which should be the required condition. Is this a correct conclusion?
Again, can I solve $y\geq 0$ to get $x\geq \frac{-b\pm {\sqrt {b^2-4ac}}}{2a}$? But then for $x$ to be real $b^2-4ac\geq 0$ must hold. So I am a little confused.
Wanted to comment, but it became too long and I think that this post will be more helpful.
We would like to find conditions such that $\forall x\in\Bbb{R}:\ ax^2+bx+c\ge 0,\quad a\ne0$.
As we know the number of roots of the function $f(x)=ax^2+bx+c$ is $0,1\text{ or }2$.
Here are examples for three functions (one for each option): $\hspace{156pt}$
Now, we want the function to be above $x$ axis or touching it, but not going below it. If there are two real solutions, the function has to go from negative values to positive values and vice verse, thus our first condition is that the function has no more than single root.
As we know, roots of function are $$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$thus, it will have at most single solution when $b^2-4ac\le 0$ (single root when $b^2-4ac=0$, no real roots when $b^2-4ac<0$).
Now, we need to check whether the function is above the axis or below it. The function has an extreme point at $\displaystyle \left(-\frac{b}{2a},\frac{4ac-b^2}{4a}\right)$. We know that $b^2-4ac\le 0$, thus the sign of $y$ coordinate depends on the sign of $a$. If $a>0$ then the point is a minima and then $y_{min}\ge 0$, thus $\forall x\in\Bbb{R}:\ f(x)\ge 0$. If $a<0$ then the point is a maxima and then $y_{max}\le 0$, thus $\forall x\in\Bbb{R}:\ f(x)\le 0$.
Conclusions: