Given $$f(x)=ax^2+bx+c\ ; \quad a\neq0.$$ Prove that it is bijective if $$x \in \Bigg[\frac{-b}{2a},\ \infty \Bigg]$$ and $$ranf=\Bigg[\frac{4ac-b^2}{4a},\ \infty \Bigg).$$
I can prove that the range of $f(x)=ax^2+bx+c$ is $ranf=\Big[\frac{4ac-b^2}{4a},\ \infty \Big)$, if $a\neq0$ and $a\gt0$ by completing the square, so I know here that the leading coefficient of the given function is positive.
I have also proved that $f(x)=ax^2+bx+c$ is injective where $f:\big[0, \infty \big)\to\Bbb R.$
But I don't know how to prove that the given function is surjective, to prove that it is also bijective.
I admit that I really don't know much in this topic and that's why I'm seeking help here. An advanced thanks to those who'll take time to help me. :)
As uniquesolution pointed out in the comments, a quadratic function cannot be surjective onto $\mathbb R$ (think of a picture of a parabola: it never reaches the $y$-values below/above its vertex). But it can be surjective onto $\left[\frac{4ac-b^2}{4a},\infty\right)$, which you seem to have already shown if you have shown that is indeed the range.