I will show a bit more, namely that the cubic polynomial in question has three real roots, the smallest of which is less than $1/4$. This is not tight: further improvements can be squeezed out of the method.

Let $u$ and $v$ be as in Iosif Pinelis's response. Then the cubic polynomial in question equals
$$p(x):=10 x^3 - 30 x^2 + (30 - 2 u) x + (2 u - v - 10),$$
and its discriminant $20(16 u^3 - 135 v^2)$ is nonnegative by [this answer of Fedor Petrov][1]. So $p(x)$ has three real roots (counted with multiplicity). Let us assume that all these roots are at least $1/4$. Then the three roots of
$$p(1/4+x)=10x^3-\frac{45}{2}x^2+\left(\frac{135}{8}-2u\right)x+\left(\frac{3u}{2}-v-\frac{135}{32}\right)$$
are nonnegative, hence in particular
$$\frac{135}{8}-2u\geq 0\geq\frac{3u}{2}-v-\frac{135}{32}.$$
It follows that
$$u\leq\frac{135}{16}\qquad\text{and}\qquad u-v-3\leq\frac{39}{32}-\frac{u}{2}.$$
Using also $(\ast)$ from my response [for this other MO question][2],
$$(u-3)^3\geq 27(u-v-3)^2\geq 27\left(\frac{u}{2}-\frac{39}{32}\right)^2.$$
Comparing the two sides, we get a contradiction to $u\leq 135/16$.


  [1]: https://mathoverflow.net/a/479088/36721
  [2]: https://mathoverflow.net/q/479107/