When $$x > 5$$

$${(x + 1)^2} + (x - 5) = {{27} \over 4}$$

$$ \Rightarrow {x^2} + 3x - 4 = {{27} \over 4}$$

$$ \Rightarrow {x^2} + 3x - {{43} \over 4} = 0$$

$$ \Rightarrow 4{x^2} + 12x - 43 = 0$$

$$x = {{ - 12 \pm \sqrt {144 + 688} } \over 8}$$

$$x = {{ - 12 \pm \sqrt {832} } \over 8} = {{ - 12 \pm 28.8} \over 8}$$

$$ = {{ - 3 + 7.2} \over 2}$$

$$ = {{ - 3 + 7.2} \over 2},{{ - 3 - 7.2} \over 2}$$

= 2.1, -5.1 [ both are rejected as x should be > 5 ]

(Therefore no solution)

For $$x \le 5$$

$${(x + 1)^2} - (x - 5) = {{27} \over 4}$$

$${x^2} + x + 6 - {{27} \over 4} = 0$$

$$4{x^2} + 4x - 3 = 0$$

$$x = {{ - 4 \pm \sqrt {16 + 48} } \over 8}$$

$$x = {{ - 4 \pm 8} \over 8} \Rightarrow x = - {{12} \over 8},{4 \over 8}$$

$$ \therefore $$ So, the equation have two real roots.