$f(x)$ and $g(x)$ are both quadratic functions with equal $x$-intercepts
$f(x)$ intercepts: $(-1,0),(0,3),(4,0)$
$g(x)$ intercepts: $(-1,0),(0,2),(4,0)$
So first I found the equation of both functions:
$f(x)=-\frac{3}{4}x^2+\frac{9}{4}x+3$
$g(x)=-\frac{1}{2}x^2+\frac{3}{2}x+2$
Then I simplified $(f\circ g)(x)=3$ to $-x^4+6x^3-7x^2-6x+8=0$
After factorizing I got $(x-1)(x-2)(x+1)(-x+4)=0$
so values of $x: 1, 2, -1, 4$
Is this solution correct? What is the simple way of solving this problem?
The simple way is to first solve $f(y) = 3$, get two solutions $y_1, y_2$, and then solve $g(x) = y_1$ and $g(x) = y_2$.
We're given directly that $y_1 = 0$ is one solution to the first equation. We can also deduce from the two other known points of $f$ the symmetry axis of $f$ ($-1$ and $4$ are equidistant form the symmetry axis), and find $y_2 = 3$ (since $3$ and $0$ must then also be equidistant form the symmetry axis).
First $g(x) = y_1$. We are given $x_1 = -1, x_2 = 4$ directly from the known points on $g$. Then $g(x) = y_2$. This time we are not given any solutions directly, so finding the explicit expression for $g$ is one way of solving this to get $x_3 = 1, x_4 = 2$.
Alternatively, since we are lucky, and $1$ and $2$ are indeed the solutions$^*$, one can use that since $g$ is a second degree polynomial, $g(x-1) - 2g(x) + g(x+1)$ has a constant (although a priori unknown) value for all $x$. Inserting $x = 0, 1, 2$ and $3$ will give you a set of equations that you can solve for $g(1)$ and $g(2)$ which shows that they are equal to $y_2$.
The intiutuve (and perhaps mentally faster) way to think about this is that as $x$ goes from $-1$ to $4$ in steps of $1$, $g(x)$ goes from $0$ to $0$ in such a way that the differences between terms follow an arithmetic sequence. We are given (from the points on $g$ which are given to us, plus the symmetry of a second degree function) $$ \begin{array}{c|ccccccccccc}x & -1 && 0&&1&&2&&3&&4\\ \hline g(x)&0&&2&&?&&?&&2&&0\\ \text{Diff}&&2&&?&&?&&?&&-2 \end{array} $$ That last line must be an arithmetic sequence, and there are three terms between $2$ and $-2$. Those three terms must therefore be $1, 0, -1$. From there you can gather that the two unknown function values must be $3$ and $3$, which are exactly the ones we are looking for.
$^*$ This might seem a bit shady, the way I've worded it. It's not. The last approach, with the table and differences was how I myself figured out that $1$ and $2$ are solutions. It worked out this time. However, had the solutions been something other than $1$ and $2$ (for instance, if we wanted to solve $g(x) = 1$ instead), then all I would've found out was that $1$ and $2$ are not solutions, and I would've had to try something else.