60. (9th grade) Solve the equation $x^{2}+|x+3|+|3-x|=4.5|x|+6$.
60. We divide the number line by the points $-3, 0$, and 3 into intervals.

If $x \leqslant -3$, then $x^{2} - x - 3 + 3 - x = -4.5 x + 6, x_{1} = -4, x_{2} = \frac{3}{2}$.

If $-3 \leqslant x \leqslant 0$, then $x^{2} + x + 3 + 3 - x = -4.5 x + 6, x_{3} = 0, x_{4} = 4.5$.

If $0 \leqslant x \leqslant 3$, then $x^{2} + x + 3 + 3 - x = 4.5 x + 6, x_{5} = 0, x_{6} = 4.5$.

If $x \geqslant 3$, then $x^{2} + x + 3 + x - 3 = 4.5 x + 6, x_{7} = 4, x_{8} = -\frac{3}{2}$.

Answer: $x_{2}, x_{4}, x_{6}, x_{8}$ are extraneous roots. $x = \pm 4 ; x = 0$ are the roots of the equation.