The equation $y = x^2 + \frac{1}{2}ax + a$ represents a parabola for all real values of $a$ [closed]
Let $a,b \in \mathbb R$ and $a \ne b$. Schow that the equation
$$ x^2 + \frac{1}{2}ax + a= x^2 + \frac{1}{2}bx + b$$
has the unique solution $x=-2.$
Hint: If $x=-2$, what is the value of $y$ if you know that the point $(x,y)$ lies on the parabola?