How to show that linear map is surjective? I have the following linear map:
$$T:\mathcal{P}^2 \to \Bbb{R}^2 $$
where $\mathcal{P}^2$ denotes the vector space of polynomials  with real coefficients having degree at most $2$
$ T $ is defined  by $$T(ax^2+bx+c)=[a+b , b−c]^t$$

I do not know how to prove that $T$ is surjective.
I know its not injective.
Yet,i do not now how to formally show that is a surjection.
I tried following the answer at How to show that a linear map is surjective?
Using the formula described:

*

*$\dim(V) = 2$,

*$\dim(\operatorname{range}T)=1$
According to the formula "will be surjective if $\dim V= \dim \operatorname{range} T$"
this map would not be surjective.
Yet, in the website i took this exercise from, it says this map is a surjective.
So, i am wondering what i did wrong, and if there are better ways to show that this map is a surjective.
I came across on this exercise on this website.

I recently make this figure to explain the concepts of injectivity and surjectivity and some related properties related to the rank and left/right inverses:

Although it is stated for matrices, it is valid for linear maps between spaces of finite dimension.
Your case corresponds to the green area in the above figure. Using the standard basis of $\mathsf{P}_2({\rm I\!R})$, that is, $\{1, x, x^2\}$, we have that your linear map is described by the matrix $M_T\in{\rm I\!R}^{2\times 3}$
$$
M_T = \begin{bmatrix}
1 & 1 & 0
\\
0 & 1 & -1
\end{bmatrix},
$$
and it easy to see that $\mathrm{rank} M_T = 2$, which proves that $M_T$ (and $T$) is surjective.
