The actual problem is to find the probability that $Ax^{2}+Bx+C=0$ has real roots. This boils down to whether or not the discriminant $B^{2}-4AC$ is non-negative. Thus, we seek $P\{B^{2}-4AC\geq 0\}$.
(Small note: $A,B,C$ are i.i.d.)
Here is what I did so far:
$$f_{A,B,C}(a,b,c) = f_{A}(a)f_{B}(b)f_{C}(c)=1 \,\,\, where \,\,\, 0<a,b,c< 1$$
Since $B^{2}-4AC\geq 0$ is equivalent to $B \geq 2\sqrt{AC}$,
$$P\{B^{2}-4AC\geq 0\} = \int_{0}^{1} \int_{0}^{1} \int_{2\sqrt{ac}}^{1} f_{A,B,C}(a,b,c) \,db\,da\,dc$$
$$P\{B^{2}-4AC\geq 0\} = \int_{0}^{1} \int_{0}^{1} (1-2\sqrt{ac}) \,da\,dc = \frac{1}{9}$$
I evaluated the last part with wolfram alpha. However, my solution is different from another source, so I am not sure where I went wrong. Any insight would be highly appreciated!
Let $\mathscr{I}$ be the integral $\displaystyle \int_0^1\int_0^1\max(1-2\sqrt{ac},0)dadc$.
You can evaluate it by variable subsitutions:
$$\begin{cases}a &= \lambda\mu\\ c &= \lambda/\mu\end{cases} \quad\longleftrightarrow\quad \begin{cases}\lambda &= \sqrt{ac}\\ \mu &= \sqrt{a/c}\end{cases}$$
The area element becomes $$da dc = \left|\begin{matrix}\mu & \lambda\\ \frac{1}{\mu}&-\frac{\lambda}{\mu^2}\end{matrix} \right| d\lambda d\mu = 2\frac{\lambda}{\mu} d\lambda d\mu$$ and the domain of integration is given by: $$ 1 - 2\sqrt{ac} \ge 0 \quad\to\quad \lambda \le \frac12\quad\text{ and }\quad a \le 1, c \le 1 \quad\to\quad \lambda \le \mu \le \frac{1}{\lambda} $$ This give us $$\begin{align} \mathscr{I} & = \int_0^{\frac12} 2\lambda ( 1-2\lambda )\left(\int_\lambda^{\frac{1}{\lambda}}\frac{d\mu}{\mu}\right) d\lambda = -4\int_0^\frac12 \lambda(1-2\lambda)\log\lambda d\lambda\\ & = -4 \left[\frac{x^2}{36}(8x - 6(4x-3)\log x-9)\right]_0^\frac12 = \frac{1}{36}(5 + 6\log 2) \sim 0.25441341898221 \end{align} $$