To solve the given functional equation for all functions $f: \mathbb{R} \to \mathbb{R}$:

$$(f(x) + f(z))(f(y) + f(t)) = f(xy - zt) + f(xt + yz),$$

we start by analyzing specific cases to deduce possible forms for $f(x)$.

1. Testing the Zero Function:

Substitute $f(x) = 0$ for all $x$. The equation becomes:

$$(0 + 0)(0 + 0) = 0 + 0,$$
which holds for all $x, y, z, t$. Thus, $f(x) = 0$ is a solution.

2. Testing the Constant Function:

Assume $f(x) = c$, where $c$ is a constant. Substituting into the equation gives:

$$(c + c)(c + c) = c + c,$$

$$4c^2 = 2c.$$

Solving $4c^2 = 2c$ yields $c = 0$ or $c = \frac{1}{2}$. Therefore, $f(x) = \frac{1}{2}$ is another solution.

3. Assuming Polynomial Form:

To explore non-constant solutions, assume $f(x)$ takes a polynomial form. Given the symmetry and the construction of the equation, check $f(x) = x^2$:

Substituting $f(x) = x^2$ leads to:

$$(x^2 + z^2)(y^2 + t^2) = (xy - zt)^2 + (xt + yz)^2.$$

Expanding both sides:

- Left side: $(x^2 + z^2)(y^2 + t^2) = x^2y^2 + x^2t^2 + z^2y^2 + z^2t^2$,
- Right side expand completely:

$$(xy - zt)^2 = x^2y^2 - 2xyzt + z^2t^2$$
$$(xt + yz)^2 = x^2t^2 + 2xyzt + y^2z^2$$

Their sum:
$$x^2y^2 + x^2t^2 + z^2t^2 + y^2z^2.$$

The expanded form matches perfectly, hence $f(x) = x^2$ is a valid solution.

Based on these analyses, the complete set of solutions is:

$$\boxed{f(x) = 0, \quad f(x) = \frac{1}{2}, \quad f(x) = x^2.}$$