In general how to solve this equation:(with Hensel's lemma and without it) $$2x^3+7x-4\equiv0 \pmod{25} $$
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$\begingroup$Without Hensel's lemma: First solve $2x^3+7x-4\equiv 0\pmod 5$, getting a solution $x_0$.
Then write $x=x_0+5k$ and substitute into the equation:
$$2(x_0+5k)^3+7(x_0+5k)-4\equiv 0\pmod {25}$$
solving for $k$. This turns out to be linear in $k$ because the high powers of $k$ have coefficients that are divisible by $25$.
With Hensel's lemma: Hensel's lemma is a shorter version of this, and can be written much like Newton's method:
$$p(x)=2x^3+7x-4$$
Let $x_0$ be a solution of $p(x)\equiv 0 \pmod 5$.
Then $$x_1=x_0 + \frac{p(x_0)}{p'(x_0)}$$
Where by the fraction, we really mean multiplication by an inverse of $p'(x_0)$ modulo $5$. Clearly, that's only going to be possible of $p'(x_0)\not\equiv 0\pmod 5$.
Then $x_1$ is a solution $\pmod {25}$.
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