$P\{B^{2}-4AC\geq 0\}$ where $A,B,C \sim U(0,1)$? 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!
 A: 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
$$\begin{align}
\mathscr{I} 
& = \int_0^{\frac12} 2\lambda ( 1-2\lambda )\left(\int_\lambda^{\frac{1}{\lambda}}\frac{d\mu}{\mu}\right) d\lambda
  = -4\int_0^\frac12 \lambda(1-2\lambda)\log\lambda d\lambda\\
& = -4 \left[\frac{x^2}{36}(8x - 6(4x-3)\log x-9)\right]_0^\frac12
  = \frac{1}{36}(5 + 6\log 2) \sim 0.25441341898221
\end{align}
$$
A: The answer above is really cool, but for those who (like me) didn't understand it at first, dlaser's second comment is almost there using conventional integration.
$$
b \ge 2\sqrt {ac}  \to 0 \le c \le \frac{1}{4},0<a<1 \cup \frac{1}{4} < c < 1, 0 < a \le \frac{1}{{4c}}
$$
Therefore, you have to add two integrals:
$$
\int\limits_0^{1/4} {\int\limits_0^1 {\int\limits_{2\sqrt {ab} }^1 {da}\, db}\, dc}  + \int\limits_{1/4}^1 {\int\limits_0^{1/4c} {\int\limits_{2\sqrt {ab} }^1 {da}\, db}\, dc}=\frac{5}{36}+\frac{\ln{2}}{6}=\frac{5+6\ln{2}}{36}
$$
A: There's another, even easier way to answer the question: condition on B. Thus:
$$
P(4AC<B^2)=\int_0^1 {P(4AC<B^2|B=b)f_B(b)\, db}
$$
Since B is uniformly distributed between 0 and 1, $f_B(b)=1$, so with some rearrangement, the above becomes
$$
\int_0^1 {P(AC<\frac{b^2}{4})\, db}
$$
and
$$
P(AC<t)=t+t\int_t^1{\frac{1}{c}\, dc}=t-t\ln{t}
$$
So
$$
\int_0^1 {P(AC<\frac{b^2}{4})\, db}=\int_0^1 {\frac{b^2}{4}-\frac{b^2}{4}\ln{\frac{b^2}{4}}\,db}
$$
The second term requires integration by parts, but the answer is correct: $\frac{5+6\ln{2}}{36}$
