The number of the real roots of the equation $$(x + 1)^2 + |x - 5| = \frac{27}{4}$$ is ______.

We need to solve $$(x + 1)^2 + |x - 5| = \frac{27}{4}$$. We consider two cases based on the absolute value.

When $$x \geq 5$$, the equation becomes $$(x + 1)^2 + (x - 5) = \frac{27}{4}$$, which simplifies to $$x^2 + 2x + 1 + x - 5 = \frac{27}{4}$$, or $$x^2 + 3x - 4 = \frac{27}{4}$$, giving $$x^2 + 3x - \frac{43}{4} = 0$$. Using the quadratic formula, $$x = \frac{-3 \pm \sqrt{9 + 43}}{2} = \frac{-3 \pm \sqrt{52}}{2}$$. Since $$\sqrt{52} \approx 7.21$$, the positive root is $$x \approx 2.1$$, which does not satisfy $$x \geq 5$$. So there are no solutions in this case.

When $$x < 5$$, the equation becomes $$(x + 1)^2 + (5 - x) = \frac{27}{4}$$, which simplifies to $$x^2 + 2x + 1 + 5 - x = \frac{27}{4}$$, or $$x^2 + x + 6 = \frac{27}{4}$$, giving $$x^2 + x - \frac{3}{4} = 0$$. Multiplying by 4: $$4x^2 + 4x - 3 = 0$$, so $$(2x + 3)(2x - 1) = 0$$. This gives $$x = -\frac{3}{2}$$ or $$x = \frac{1}{2}$$. Both values satisfy $$x < 5$$, so both are valid solutions.

Therefore, the number of real roots is $$2$$.