Solve the congruence $x^3+2x-3\equiv{0}\pmod{45}$ Solve (if possible)the congruence involving polynomial 
$x^3+2x-3\equiv{0}\pmod{45}$

My work:
Since $45=3^2\cdot5$, we have 
$x^3+2x-3\equiv{0}\mod(3)$ and $x^3+2x-3\equiv{0}\pmod 5$
In $\mathbb{Z}_3$,
$x^3+2x-3=(x-1)(x^2+x+3)\equiv{0}\pmod 3$
We have $[0],[1],[2]$
In $\mathbb{Z}_5$,
$x^3+2x-3=(x-1)(x^2+x+3)\equiv{0}\pmod 5$
We have $[0],[1],[2],[3],[4]$
So does it mean I have 3 solutions, $[0],[1],[2]$???
And how should I finish it up?

Here is one way to find the roots $\rm\,mod\ 9.\:$ Since $\rm\,9\mid(x-1)(x^2+x+3),\:$ by unique factorization, either $\rm\,9\mid x-1,\,\ 9\mid x^2+x+3,\:$ or $\rm\:3\mid x-1,\,x^2+x+3.\:$ The last case  is impossible since $\rm\:3\mid x-1\,\Rightarrow\,mod\ 3\!:\ x\equiv 1\,\Rightarrow\, x^2+x+3\equiv 2\not\equiv 0.\:$ In the second case we have $\rm\:mod\ 9\!:\ x^2+x+3\equiv x^2+x-6 = (x-2)(x+3).\:$ Again, either $\rm\,9\mid x-2,\,\ 9\mid x+3,\,$ or $\rm\,3\mid x-2,\,x+3.\:$ The latter is impossible, else $\rm\,2\equiv -3\,\ (mod\ 3).\:$ So this case yields two roots $\rm\:x\equiv 2,\, -3.\:$  Finally, the first case $\rm\,9\mid x-1\,$ yields the root $\rm\:x\equiv 1\,\ (mod\ 9).$
$\rm mod\ 5\!:\ x^2+x+3\equiv x^2+x-2 \equiv (x-1)(x+2),\:$ so there are two roots $\rm\:x\equiv 1,-2\,\ (mod\ 5).$
Now employ the CRT (Chinese Remainder Theorem) to combine the solutions modulo $9$ and $5$ to obtain all of the solutions modulo $\,\rm lcm(9,5) = 45.$
