Prove that $ ax^2+bx+c=0 $ has at least one root in $(0,1)$ if $10a+12b+15c=0$

Consider $x = \frac{10}{12}$:

$$\begin{align*} f\left(\frac{10}{12}\right) &= a\left(\frac{10}{12}\right)^2+b\left(\frac{10}{12}\right)+c\\ &= \frac{10}{144}\left(10a+12b+\frac{144}{10}c\right)\\ &= \frac{10}{144}\left(10a+12b+15c-\frac{6}{10}c\right)\\ &= -\frac{6}{144}c \end{align*}$$

And consider $x=0$:

$$f(0) = 0^2a + 0b + c = c$$

If $c\ne 0$, by intermediate value theorem, there must be an $x\in\left(0, \frac{10}{12}\right)\subset(0,1)$ which satisfies $f(x)= 0$.

If $c=0$, $\frac{10}{12}\in(0,1)$ is a root.

How I obtained $x = \frac{10}{12}$:

I assume $x = 12k$ and $x^2 = 10k$ for some real number $k$. By solving $10k = (12k)^2$, $k$ is chosen to be the non-zero root $k = \frac{10}{144}$, which appears as the multiplier above. And $x = 12k = \frac{10}{12}$.