Solving quadratic equations in the field $F_5$ Let $y = x^2 + 2x + 2 = 0$.
Solve the equation in the field $F_5$.
So I used the common $b^2 - 4ac$ formula and got that $x$ is either $-1/2$ or $-3/2$ but I'm not sure if this is in the field...

We have $a=1$, $b=2$, $c=2$, so $b^2-4ac=4-8=-4=1$. Moreover $2^{-1}=3$, so
$$
x=2^{-1}(-2\pm1)=3(-2\pm1)
$$
which means $x=3(-3}=-9=1$ or $x=3(-1)=-3=2$. The formula
$$
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
$$
indeed means
$$
x=(2a)^{-1}(-b\pm\sqrt{b^2-4ac})
$$
Check:
$$
1^2+2\cdot1+1=0,\qquad 2^2+2\cdot2+2=0
$$
