An equation of form $x^{2}+ax+b=0$ might have infinite amount of solutions in a ring $(R,+,\cdot)$ An equation of form $x^{2}+ax+b=0$ might have infinite amount of solutions in a ring $(R,+,\cdot)$.
Now I am a bit lost here.
The definition for ring is that $(R,+)$ is Abel and $(R,\cdot)$ is a monoid.
I just wonder what in earth they are after in this exercise?
I should find a equation of that form and show that it has infinite amount of solutions. But it feels just a bit absurd.

After receiving these very good answers. I feel that I should write an example.
Let's look at matrix ring $(M_{2}(\mathbb{Z}/4\mathbb{Z}),+,\cdot)$, which has the usual matrix addition and multiplication. Now, when $n>1$, the $n \times n$ matrix is not commutative. Now we can calculate that
$$
\begin{bmatrix}
2 & 0\\
0 & 0
\end{bmatrix}\cdot
\begin{bmatrix}
2 & 0\\
0 & 0
\end{bmatrix}
=
\begin{bmatrix}
4 & 0\\
0 & 0
\end{bmatrix}
=
\begin{bmatrix}
0 & 0\\
0 & 0
\end{bmatrix}
$$
so we have solution to a equation
$$
X^{2}=0.
$$
We can find an example of infinite ring that has infinitely many solutions to the equation above.
For example such is matrix ring $(M_{2}(\mathbb{R}),+,\cdot)$ where infinitely many solutions can be found using matrix of form
$$
\begin{bmatrix}
0 & a\\
0 & 0
\end{bmatrix}
$$
where $a\in\mathbb{R}$.
 A: Example
$$x^2+1=0\;,\;\;x\in\Bbb H=\text{Hamilton's Quaternions}$$
A: HINT: Let $R$ be the ring of infinite sequences of zeroes and ones with coordinatewise addition and multiplication modulo $2$.
A: Remember that a ring may have zero divisors. 
For example, if $M_2(R)$ is a $2\times 2$ matrices ring, over any non-zero ring $R$ including a nilpotent element of the form $a^2=0$, then, even the equation 
$$
x^2=\mathbf{0}=
\begin{bmatrix}
0 & 0 \\
0 & 0
\end{bmatrix}
$$ 
has an infinite number of solutions. In fact, any matrix 
$$
x=
c\begin{bmatrix}
a & 0 \\
0 & 0
\end{bmatrix}=
\begin{bmatrix}
ca & 0 \\
0 & 0
\end{bmatrix}
$$
for any $c\in R$, is a solution of the above. 
P.S.: If you want a concrete example of the above idea, take any infinite ring $R$ with zero divisors $b,d$, such that $bd=0$ and $db\neq 0$ (matrix rings provide a host of such examples). Then the element $a=db\neq 0$ is nilpotent, since $a^2=(db)^2=d(bd)b=0$. Now, your required ring will be $M_2(R)$ and inside it, the equation $x^2=\mathbf{0}$, will have an infinite number of solutions of the above form. 
An even simpler example has already been mentioned in an update of OP: Take $R=\mathbb{R}$. Then the equation $x^2=\mathbf{0}$ will have an infinity of solutions inside the ring $M_2(\mathbb{R})$, of the form: 
$$
\begin{bmatrix}
0 & a\\
0 & 0
\end{bmatrix}
$$
for any $a\in\mathbb{R}$. 
A: Consider the ring $R=\mathbb{Z}\times \mathbb{Z}$ and the equation $(a,0)x=(0,0)$ with $a \in \mathbb{Z}$ over $R$. It has infinitely many solutions of the form $\{(0,n):n \in \mathbb{Z}\}$. Try to use this idea.
A: Summarizing a bit all the answers here: If $R$ is an integral domain (i.e. it is commutative and does not have divisors of $0$), then you may consider its field of fractions $Q(R)$ and then embed this one in its algebraic closure $\overline {Q(R)}$. Since $\overline {Q(R)}$ is commutative and algebraically closed, any $2$nd degree equation will have at most $2$ distinct roots. Since $R \subseteq \overline {Q(R)}$, it follows that any equation of the form $x^2 + ax + b = 0$ with $a,b \in R$ will also have at most $2$ distinct roots in $R$, clearly not what you want.
In order to avoid the conclusion obtained above, it follows that you must look for a ring which is not an integral domain. There are two possible approaches: either look for a non-commutative ring (rings of matrices, for instance), or for a ring with divisors of $0$ (products of integral domains, rings of matrices).
