I will show a bit more, namely that the cubic polynomial in question has three real roots, the smallest of which is less than $9/25$. 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 $9/25$. Then the three roots of
$$p(9/25+x)=10x^3-\frac{96}{5}x^2+\left(\frac{1536}{125}-2u\right)x+\left(\frac{32u}{25}-v-\frac{8192}{3125}\right)$$
are nonnegative, hence in particular
$$\frac{1536}{125}-2u\geq 0\geq\frac{32u}{25}-v-\frac{8192}{3125}.$$
It follows that
$$u\leq\frac{768}{125}\qquad\text{and}\qquad u-v\leq \frac{8192}{3125}-\frac{7u}{25}.$$
However, my response [for this other MO question][2] also shows that 
$$u\geq 3\qquad\text{and}\qquad |u-v-3|\leq\left(\frac{u}{3}-1\right)^{3/2},$$
hence in fact
$$3-\left(\frac{131}{125}\right)^{3/2}\leq u-v\leq\frac{5567}{3125}.$$
This is a contradiction, because the left-hand side exceeds the right-hand side.


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