The number of real solutions of the equation, $$x^2 - |x| - 12 = 0$$ is:

We have to solve the equation $$x^{2}-|x|-12=0$$ and count how many real values of $$x$$ satisfy it.

The expression contains the absolute value $$|x|$$, so we split the work into two cases depending on whether $$x$$ is non-negative or non-positive. Recall the definition $$|x|=\begin{cases}x,&x\ge 0\\-x,&x\le 0\end{cases}.$$

Case 1 : $$x\ge 0$$. Here $$|x|=x$$, so substituting this into the equation gives

$$x^{2}-x-12=0.$$

This is a quadratic equation of the standard form $$ax^{2}+bx+c=0$$ with $$a=1,\;b=-1,\;c=-12$$. The quadratic formula states $$x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}.$$ Applying it,

$$x=\dfrac{-(-1)\pm\sqrt{(-1)^{2}-4(1)(-12)}}{2(1)} =\dfrac{1\pm\sqrt{1+48}}{2} =\dfrac{1\pm 7}{2}.$$

So the two roots are

$$x=\dfrac{1+7}{2}=4\quad\text{and}\quad x=\dfrac{1-7}{2}=-3.$$

Because we are inside the case $$x\ge 0$$, only $$x=4$$ is admissible; $$x=-3$$ must be discarded.

Case 2 : $$x\le 0$$. Now $$|x|=-x$$, and the equation becomes

$$x^{2}+x-12=0.$$

Again using the quadratic formula with $$a=1,\;b=1,\;c=-12$$, we obtain

$$x=\dfrac{-1\pm\sqrt{1^{2}-4(1)(-12)}}{2(1)} =\dfrac{-1\pm\sqrt{1+48}}{2} =\dfrac{-1\pm 7}{2}.$$

Thus the roots are

$$x=\dfrac{-1+7}{2}=3\quad\text{and}\quad x=\dfrac{-1-7}{2}=-4.$$

But we must respect the condition $$x\le 0$$ for this case, so only $$x=-4$$ is acceptable; $$x=3$$ is rejected.

Gathering results from both cases, the values that satisfy the original equation are

$$x=4\quad\text{and}\quad x=-4.$$

There are exactly two real solutions.

Hence, the correct answer is Option A.