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**compNewBe** I need to prove that $F[x]/\langle p(x)\rangle$ contains both roots of $p(x)$ where p is irreducible and has degree 2

Here is my proof but I dont think I am going at it the right way

Suppose p(x) is irreducible and has degree 2. We want to prove that $F[x]/\langle p(x)\rangle$ contains both roots of p(x). Since p(x) has degree 2 it is of the form $ax^2 + bx + c$. Thus the roots of $ax^2 + bx + c$ are $\frac{-b + \sqrt{b^2 -4ac}}{2a}$ and $\frac{-b + \sqrt{b^2 -4ac}}{2a}$. Since though p(x) is a generator $a,b,2 \in F$[x]. All we need to do now is prove that $-b + \sqrt{b^2 -4ac}$ and $-b - \sqrt{b^2 -4ac}$ are in $F[x]/\langle p(x)\rangle$. This is obtained by multiplying by 2a and subtracting $\frac{-b}{2a}$ But since F is a field all we need to prove is that $-b + \sqrt{b^2 -4ac}$ is in F because F is also abelian group.

If i am on the right path I cannot think of a way to prove the rest any help would be great