The number of real roots of the equation $x|x|-5|x+2|+6=0$, is :

$$ \begin{aligned} & x|x|-5|x+2|+6=0 \\\\ & \text { Case-I : } \\\\ & \text { When } x<-2 \text { then } \\\\ & -x^2+5(x+2)+6=0 \\\\ & \Rightarrow x^2-5 x-16=0 \\\\ & \Rightarrow x=\frac{5 \pm \sqrt{25+64}}{2} \\\\ & \therefore x=\frac{5-\sqrt{89}}{2} \text { is accepted } \end{aligned} $$ <br/><br/><b>Case-II :</b> <br/><br/>$$ \begin{aligned} & \text { When } -2 \leq x < 0 \text { then } \\\\ & -x^2-5(x+2)+6=0 \\\\ \Rightarrow & x^2+5 x+4=0 \\\\ \Rightarrow & (x+1)(x+4)=0 \\\\ & x=-1 \text { is accepted } \end{aligned} $$ <br/><br/><b>Case-III :</b> <br/><br/>$$ \begin{aligned} & \text { When } x \geq 0 \text { then } \\\\ & x^2-5(x+2)+6=0 \\\\ \Rightarrow & x^2-5 x-4=0 \\\\ & x=\frac{5 \pm \sqrt{25+16}}{2} \\\\ & =\frac{5 \pm \sqrt{41}}{2} \\\\ & x=\frac{5 - \sqrt{41}}{2} \text { is accepted } \\\\ \therefore & 3 \text { real roots are possible. } \end{aligned} $$