Absolute value 1

$\left| { x }^{ 2 }-5x+6 \right| +\left| x-1 \right| =5\\ \left| (x-3)(x-2) \right| +\left| x-1 \right| =5\\ So\quad we\quad now\quad have\quad 4\quad cases,\\ x\ge 3,\quad 2\ge x\ge 3,\quad 1\ge x\ge 2,\quad x\le 1\\ Solving\quad for\quad these\quad domains\quad we\quad \\ get\quad that\quad x=4\quad and\quad x=3-\sqrt { 7 } \\ Hence\quad only\quad 2\quad solutions\quad are\quad possible.\quad \\$