Finding parameters for a quotient of a polynomial ring 
Let $a,b \in \mathbb{R}, T_{a,b} := \mathbb{R}[x] \ /\langle x^2+ax+b\rangle$, where $\langle x^2+ax+b\rangle$ is the ideal generated by $x^2+ax+b$.
1) for which $a,b \in \mathbb{R}$ is $T_{a,b}$ a field?
2) for which $a,b \in \mathbb{R}$ is $x \in T_{a,b}$ invertible ($x$ refers to the $x$ in $\mathbb{R}[x]$), and how does the inverse look like? 

1) 
Let $b = 0 \Rightarrow T_{a,0} = \mathbb{R}[x] \ /\langle x^2+ax\rangle$.
Let $u = x, v = a x \Rightarrow u,v \in T_{a,0}$
$uv = x^2+ax \not\in T_{a,0} \Rightarrow T_{a,0}$ not field $\Rightarrow T_{a,b}$ field $\Rightarrow b \neq 0$.
This is all that I saw. Can you please tell me how the other conditions of a,b can be found, so that $T_{a,b}$ field applies? I tried to identify all roots of $x^2 + ax + b$, but didn't find an approach that helped me to go on.
2)
I tried to somehow utilize that $x^2 + ax + b = 0$ for $T_{a,b}$, but got nowhere. Can you please help me to find a solution?

2) We  have 
$$
x(x+a)=x^2+ax=-(ax+b)+ax=-b.
$$
Thus 
$$
\frac{1}{x}=\frac{x+a}{x(x+a)}=\frac{x+a}{-b}
$$
