Given,
x∣x∣−5∣x+2∣+6=0
Now taking case (1) when x<−2 we get,
−x2+5x+10+6=0
⇒x2−5x−16=0
⇒x=25−89, ignoring x=25+89 as x<−2, so one solution is possible here,
Now taking case (2) when −2≤x<0 we get,
−x2−5x−10+6=0
⇒x2+5x+4=0
⇒x=−1, ignoring x=−4 as −2≤x<0
Now taking case (3), when x≥0 we get,
x2−5x−10+6=0
⇒x2−5x−4=0
⇒x=25+41, ignoring x=25−41 as x≥0
So, from all cases we get total 3 solutions.