I will show a bit more, namely that the cubic polynomial in question has three real roots, the smallest of which is at most $$u:=\frac{5-\sqrt{15}}{5}=0.22540333\dotsc$$
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. So $p(x)$ has three real roots (counted with multiplicity). Let us assume that all these roots exceed $d$. Then the three roots of $$p(d+x)=10x^3-6\sqrt{15}x^2+2\left(9-u\right)x+\left(2\sqrt{\frac{3}{5}}u-v-6\sqrt{\frac{3}{5}}\right)$$ are positive, hence in particular $$u<9\qquad\text{and}\qquad 2\sqrt{\frac{3}{5}}(u-3)<v.$$ Using also $(\ast)$ from my response for this other MO question, $$(u-3)^3\geq 27(u-v-3)^2>27\left(2\sqrt{\frac{3}{5}}-1\right)^2(u-3)^2.$$ Comparing the two sides, we obtain a contradiction: $$6>u-3>27\left(2\sqrt{\frac{3}{5}}-1\right)^2>8.$$