<s>[INST] Probability of $ax^2 + bx + c = 0$ having real solutions 
$a$, $b$, $c$ are random integer numbers between $1$ and $100$ (including $1$ and $100$, and uniformly distributed). What is the probability that the equation $ax^2 + bx + c = 0$ has real solutions?

This is from a final high school math exam, and I don't know how to get the answer without using computer.
 [/INST] To get a real solution, the discriminant $b^2-4ac\geq0$, so $b\geq 2 \sqrt{ac}$, and $\frac{b^2}{4}\geq ac$. All of our variables are real so there is no need to consider negatives.
Values of $a$ and $b$ can range from $1$ to $n$. The probability of each separate combination is $\frac{1}{n}*\frac{1}{n}$. Considering each combination of $a$ and $b$ separately:
The number of values for $c$ that will give a real solution is $\bigg\lfloor \lfloor \frac{b^2}{4} \rfloor / a \bigg\rfloor$ (restricted to less than 100), so the probability that a random $c$ will give a real solution is $\bigg\lfloor \lfloor \frac{b^2}{4} \rfloor / a \bigg\rfloor * \frac{1}{n}$.
The probability that a random $a$ and random $c$ will give a real solution is $\bigg\lfloor \lfloor \frac{b^2}{4} \rfloor / a \bigg\rfloor * \frac{1}{n} * \frac{1}{n}$.
So, for each possible value of $b$, the probability of a real solution existing is $\frac{1}{n^2}\sum_{a=1}^{n} \bigg\lfloor \lfloor \frac{b^2}{4} \rfloor / a \bigg\rfloor$.
Then, for all possible values of b, the probability of a real solution existing is $\frac{1}{n^3}\sum_{b=1}^{100}\sum_{a=1}^{100} \bigg\lfloor \lfloor \frac{b^2}{4} \rfloor / a \bigg\rfloor$.
I calculated this in Excel for $n=100$ and also got $0.249222$ as did several commenters. I do not see how one would be expected to do this by hand. There is no immediately obvious pattern. Perhaps trying smaller $n$ might offer insight.
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