Let F be a field of characteristic 2. Find the roots of $x^2+bx+c$ if we know $d$ such that $d(d+1)=b^{-2}c$. 

Let $F$ be a field of characteristic 2. Find the roots of $x^2+bx+c$ if we know $d$ such that $d(d+1)=b^{-2}c$, where ,b,c,d $\in F$. 


I tried using the formula $\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-b\pm \sqrt{b^2-4c}}{2}$, and I am stumped by the fact that $2\equiv0 \mod 2$, so there will be division by $0$. 
I also considered another method: To let $x+i$ and $x+j$ be the factor of $x^2+bx+c$. Then, $(x+i)(x+j)=x^2+(i+j)x+ij$, which implies $i+j=b$ and $ij=c$. But completely stumped as well?
How should I attempt this question? I have basic elementary group theory algebra knowledge.

If you multiply by $b^{-2}$ you get that $x^2+bx+c=0$ is equivalent to $(b^{-1}x)^2+b^{-1}x+d(d+1)=0$, which is equivalent to $b^{-1}x(b^{-1}x+1)=d(d+1)$.  It therefore reduces to solving $y(y+1)=d(d+1)$ with $y=b^{-1}x$.  Those solutions are $y=d$ and $y=d+1$, hence the original solutions are $x=bd$ and $x=b(d+1)$.
