$$\begin{gathered} Number of integeral values of x for which \\ \left|\left|\left|x^2-x+4\right|-2\right|-3\right| = x^2+x-12 is \end{gathered}$$

$$\begin{gathered} \left|\left|\left|x^2-x+4\right|-2\right|-3\right| = x^2+x-12 \\ \Rightarrow \left|\left|x^2-x+2\right|-3\right|= x^2+x-12 \end{gathered}$$

$$\begin{gathered} \Rightarrow \left|x^2-x-1\right|= x^2+x-12 \\ \Rightarrow -( x^2-x-1)= x^2+x-12 or (x^2-x-1)=x^2+x-12 \\ \Rightarrow 2x^2=-13 or 2x=11 \end{gathered}$$

$$\begin{gathered} not possible \therefore x=\frac{11}{2} \\ hence only one solution \end{gathered}$$