How many different values of a for which the system of equations
$\displaystyle y^2 + y + x^2 = 2$ and $\displaystyle x + y = a$ has only one solution?
From the second equation, we have $\displaystyle \displaystyle \begin{align*} x = a - y \end{align*}$. Substituting into the first equation gives
$\displaystyle \displaystyle \begin{align*} y^2 + y + \left(a - y\right)^2 &= 2 \\ y^2 + y + a^2 - 2a\,y + y^2 &= 2 \\ 2y^2 + \left(1 - 2a\right)y + a^2 - 2 &= 0 \end{align*}$
For there to only be one solution, the discriminant must be 0, so
$\displaystyle \displaystyle \begin{align*} \left(1 - 2a\right)^2 - 4(2)\left(a^2 - 2\right) &= 0 \\ 1 - 4a + 4a^2 - 8a^2 + 16 &= 0 \\ -4a^2 - 4a + 17 &= 0 \end{align*}$
Checking the discriminant of this quadratic tells us how many values of a satisfy this equation, and therefore tell us how many values of a will give one solution to the original set of equations...
$\displaystyle \displaystyle \begin{align*} (-4)^2 - 4(-4)(17) &= 16 + 272 \\ &= 288 \\ &> 0 \end{align*}$
Since this is positive, there are two values of a which will satisfy the set of equations.