Construct a field of 25 elements.
Solution 1:
Yes, you can take $x^2+2$: any degree two irreducible polynomial would do.
If you take $K=\mathbb{F}_p[x]/\langle x^2+2\rangle$ and denote by $u$ an element such that $u^2=3$, you can find in it a root of $x^2+x+1$: $$ (a+bu)^2+(a+bu)+1=0 $$ is the same as $$ a^2+3b^2+a+1=0,\qquad 2ab+b=0 $$ so we must take $a=-1/2=2$ and so $3b^2=3$; thus $2+u$ and $2-u$ are the solutions.