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

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(2) : Case-I : When $x \leq 5$ In this case, we have $(x+1)^{2}-(x-5)=\frac{27}{4}$\ $\Rightarrow \quad(x+1)^{2}-(x+1)-\frac{3}{4}=0$\ $\Rightarrow \quad x+1=\frac{3}{2},-\frac{1}{2}$\ $\Rightarrow \quad x=\frac{1}{2}, \frac{-3}{2}$ which are real Case-II: When $x \geq 5$\ In this case, we have,\ $(x+1)^{2}+(x-5)=\frac{27}{4}$\ $\Rightarrow(x+1)^{2}+(x+1)-\frac{51}{4}=0$\ $\Rightarrow x+1=\frac{-1 \pm \sqrt{52}}{2}$,\ which are not possible as $x \geq 5$\ $	herefore \quad$ The equation has 2 real roots.\