Let $b_A(x)$ and $b_B(x)$ denote the bids that players $A$ and $B$ make, respectively, when their valuation is $x$. We can assume that these are strictly monotonically increasing functions (and later check for consistency). Then the expected profit of player $A$ with valuation $x$ is
$$
\frac{b_B^{-1}(b_A(x))-10}{30-10}(x-b_A(x))\;.
$$
Multiplying by $20$ and setting the derivative with respect to $b_A(x)$ to zero yields
$$
{b_B^{-1}}'(b_A(x))(x-b_A(x))-\left(b_B^{-1}(b_A(x))-10\right)=0\;.
$$
In equilibrium, $b_B\equiv b_A\equiv b$, which yields
$$
\frac{x-b(x)}{b'(x)}-(x-10)=0\;,
$$
or
$$
b'(x)+\frac{b(x)}{x-10}=\frac x{x-10}\;.
$$
The general solution of the homogeneous equation is $b(x)=\frac c{x-10}$. The ansatz $b(x)=mx+n$ yields the particular solution $b(x)=\frac x2+5$ of the inhomogeneous equation, so the general solution is $b(x)=\frac x2+5+\frac c{x-10}$. Since $b(x)$ has a pole at $x=10$ for $c\ne0$, we choose $c=0$ to obtain $b(x)=\frac x2+5$. If player $B$ uses this strategy and player $A$ bids $b_A(x)$, the expected profit of player $A$ is
$$
\frac{2b_A(x)-20}{30-10}(x-b_A(x))\;,
$$
which is quadratic in $b_A(x)$ with negative coefficient. Thus the stationary point found above is a global maximum, and the strategy profile $(b,b)$ with $b(x)=\frac x2+5$ is indeed a Nash equilibrium.