The number of irreducible monic polynomials of degree $n$ (only when $n$ is prime) over the field of characteristic $p$ is $\frac{p^n-p}{n}$. In your case $p=5$ and $n=2$
Alternately Gauss gave the following result,
The number of irreducible monic polynomials of
degree $n$ over $F_q$ is given by $$N_q(n)=\frac{1}{n}\sum_{d|n}\mu(d)q^{n/d}$$ where $\mu$ is the Mobius function.
Note: As this result is given in this answer.
The number of irreducible monic polynomials of degree $n$ (only when $n$ is prime) over the field of characteristic $p$ is $\frac{p^n-p}{n}$. In your case $p=5$ and $n=2$
Alternately Gauss gave the following result,
The number of irreducible monic polynomials of degree $n$ over $F_q$ is given by $$N_q(n)=\frac{1}{n}\sum_{d|n}\mu(d)q^{n/d}$$ where $\mu$ is the Mobius function.
For a proof see this.