Amount of solutions to an equation Briliant.org asks for how many solutions there are to this equation:
$x^3 - 8x^2 + 3x^2 + 3x^3 - 7x^2 = x + 2x^3 - 2x - 2x^3 + x$
Now if I put them in order:
$ 3x^3 + x^3 -8x^2 -7x^2 + 3x^2 = 2x^3 - 2x^3  - 2x + x + x$
and combine
$ 4x^3 -12x^2 = 0$
I could add $12x^2$ to both sides and divide by $4x^2$ so it will become
$x = 3$
Now I would say there is one solution, also khan academy told me there is either one, infinite or no solution to an equation. But according to Brilliant.org, there are 2 solutions: either 3 or 0. They use the distributive property after combining so that
$ 4x^3 -12x^2 = 0$ becomes $ 4x^2(x-3) = 0$. They conclude that either x can be 3 or 0 so one of the fractions become 0 and so will the product, using the zero product property.
It does make perfect sense since I can just put the answer in the original equations and they work... But now I am also confused, isn't this true for most equations and can equations have multiple solutions, maybe even more than 2?

A cubic equation has either $3$ real roots or $1$ real root and $2$ complex roots. Here, we have three real roots, two of which are identical.
\begin{align*}
x^3 - 8x^2 + 3x^2 + 3x^3 - 7x^2 &= x + 2x^3 - 2x - 2x^3 + x\\
4 x^3 - 12 x^2 &= 0\\
(x-0)(x-0)(x-3) & =\frac{0}{4}
\implies x\in\big\{0,0,3\big\}
\end{align*}
